Atomic Structure Chemistry

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SUBJECT – CHEMISTRY

TOPIC – ATOMIC STRUCTURE

COURSE CODE – AISM-09/C/ATS

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Contents:- Atomic Structure

Fundamental Particles..................................3

Discovery of Electron...................................4

Discovery of Proton.....................................8

Discovery of Neutron..................................12

Atomic Terms............................................13

Atomic Models...........................................17

Electronic Magnetic Radiation.......................24

Atomic Spectrum........................................28

Planck‟s Quantum Theory.............................29

Bohr‟s Atomic Model....................................32

Quantum Numbers......................................50

Pauli‟s Exclusion Principle..............................54

Shapes Ans Sizes of Orbitals..........................54

Hund‟s rule of maximum multiplicity................59

Electronic configuration of elements................60

Dual Character............................................68

Derivation of De-Broglie Equation...................69

Heisenberg‟s uncertainty principle……………………..76

Quantum mechanical model of model……………….80

Photoelectric Effect………………………………………………83

Answer to excercises…………………………………………..88

Miscellaneous excercises…………………………………….89

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FUNDAMENTAL PARTICLES

Atoms are made up-essentially, of three fundamental particles, which differ in

mass and electric charge as follows:

Electron Proton Neutron

Symbol e or e- p n

Approximate relative

mass

1/1836 1 1

Approximate relative

charge

1 +1 0

Mass in kg 9.109534 10 31 1.6726485 10 27 1.6749543 10 27

Mass in amu 5.4858026 10 4 1.007276471 1.008665012

Actual charge/C 1.6021892 10 19 1.6021892 10 19 0

The atomic mass unit (amu) is 1/12th of the mass of an individual atom of

6C12, i.e. 1.660565 10 27 kg. The neutron and proton have approximately

equal masses of 1 amu and the electron is about 1836 times lighter; its mass

can sometimes be neglected as an approximation. The electron and proton

have equal, but opposite, electric charges; the neutron is not charged.

The existence of electrons in atoms was first suggested, by J.J. Thomson, as a

result of experimental work on the conduction of electricity through gases at

low pressures, which produces cathode rays and x-rays, and a study of

radioactivity by Becquerel, the Curies and Rutherford.

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An atom is electrically neutral, and if it contains negatively charged electrons

it must also contain some positively charged particles, and the supposition

that they existed within atoms came about as a result of Rutherford's

experiments in which he bombarded elements with the - rays and

-rays were given off by radioactive elements. The neutron was discovered in

1932 by James Chadwick by bombarding beryllium with rays.

DISCOVERY OF ELECTRON:

CATHODE RAYS

During the latter half of the nineteenth century, it was found that while

normally dry gases do not conduct an electric current, they do so under very

low pressure and then patches of light are seen. The passage of electricity

through gases as studied by a number of physicists, particularly by Faraday,

Davy, Crookes and J.J. Thomson.

When a current of high voltage (10,000 volts) is passed through a gas of air

kept at a very low pressure (0.01 – 0.03 mm) blue rays are seen emerging

from the case. These rays are called “Cathode Rays”.

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Cathode

To Vacuum Pump+ (Anode)

Glass Tube

Fluorescence

Cathode Rays

CATHODE RAY EXPERIMENT (DISCHARGE TUBE)

Some of the important properties of the cathode rays studied by Sir J.J.

Thomson and others are given below:

Cathode rays come out at right angles to the surface of the cathode

and move in straight lines.

Their path is independent on the position of the anode.

They produce phosphorescence on certain salts like ZnS and

fluorescence on glass.

They blacken photographic plates.

The rays pass through thin sheet of metals. If the metal sheet is too

thick to be penetrated the rays cast a shadow.

They produce X ray when they strike a metal.

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The rays ionize a gas through which they pass.

They heat a substance on which they fall.

They rotate a light wheel placed in their paths. This shows that

cathode rays contain material particles having both mass and velocity.

The mass of a particle present in cathode rays is found to be 1/1837 of

H atom. This shows that the particle is of sub atomic nature.

Cathode rays are deflected by a magnetic or an electric field showing

the particle to be electrically charged, the direction of deflection shows

that they are negatively charged.

Cathode rays contain the smallest unit of negative charge.

No cathode ray was produced when the tube was completely

evacuated.

Different gases produce same cathode rays as they have the same

e/m (charge/mass) ratio. This indicates that the particles present in

cathode rays are fundamental constituent of all matter.

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Sir J.J. Thomson named these negatively charged sub atomic particles as

electron.

“A sub atomic particle which is a fundamental constituent of all matter

having a mass 1/1837th of a H atom and which carries the smallest unit of

negative charge is called an electron”.

Determination of Velocity and Charge/mass (e/m) ratio of Electrons:

Sir J.J. Thomson (1897) extended the cathode ray experiment for the

determination of velocity of electrons and their charge/mass ratio, The value

of e/m for an electron = 1.76 108 C/g. For the H+ ion (proton),

e/m = 96500/1.008 C/g.

Millikan’s Oil Drop Method: Determination of Charge on an Electron:

In 1909, Millikan measured the charge on an electron by his oil drop

method. In this method a spray of oil droplets is produced by an atomizer,

some of which pass through an opening into a viewing chamber, where we

can observe them with a microscope. Often these droplets have an electric

charge, which is picked up from the friction forming the oil droplets. A

droplet may have one or more additional electrons in it, giving it a negative

charge.

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As the droplet falls to the bottom of the chamber, it passes between two

electrically charged plates. The droplet can be suspended between them; we

adjust the voltage in the plates so that the electrical attraction upward just

balances the force of gravity downward. We then use the voltage needed to

establish this balance to calculate the mass - to charge ratio for the droplet.

Because we already know the mass of the droplet we can find the charge on

it.

Millikan's found that the charge on all droplets could be expressed as whole

number multiples of e, where the value of e is 1.602 10-19 C. By combining

e/m. ratio and 'e' we calculate mass of the electron

19

e 8

e 1.6022 10M

e/m 1.76 10

= 9.104 10-31 kg

This very small value shows that the electron is a subatomic particle. Thus

charge on an electron = 1.602 10–19C.

DISCOVERY OF PROTON:

POSITIVE RAYS OR CANAL RAYS

Atoms are electrically neutral. Hence after the discovery of the negatively

charged constituent (electron) of an atom, attempts were made to discover

the positively charged counterpart of electrons. By using a discharge tube

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containing a perforated cathode. Goldstein (1886) found that some rays

passed through these holes in a direction opposite to that of the cathode

rays.

Anode

To Vacuum Pump

Cathode

Positive Rays

Cathode Rays

POSITIVE RAYS

These are called the positive rays or canal rays. J.J. Thomson (1910)

measured their charge by mass ratio from which he was able to deduce that

these contain positive ions. Their properties are:

They are positively charged.

The positive charge is either equal to or whole number multiple of the

charge on an electron.

When hydrogen gas was filled in the discharge tube the positive

charge on the positive rays was equal to the negative charge on an

electron, and the mass was less than the hydrogen atom.

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Unlike cathode rays the properties of positive rays are characteristics

of the gas in the tube.

The deflection of positive rays under the influence of an electric or

magnetic field is smaller than that of the cathode rays for the same

strength of field. This shows that the positive rays have a greater mass

than that of electrons.

The mass of the positive rays depends on the atomic weights or

molecular weights of the gases in the discharge tube. The charge/mass

ratio also varies because the change in positive charge on the rays. It

may be either equal to or integral multiple of the charge on an

electron.

The lightest of all particles identified in positive rays from different

elements was one with a mass very slightly less than that of hydrogen

atom (or nearly equal to H atom). The lightest positively charged

particle is called a proton (P or P+). Positive rays are atomic or

molecular resides from which some electrons have been removed. The

removed electrons constitute the cathode rays and the positive

residues form the positive or canal rays.

Positive Rays Cathode Rays

H H+ e–

O O+ e–

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O2 O2+ e–

O2 O22+ 2e–

The mass of a proton is very slightly less than that of a H atom. This shows

that protons are sub atomic particles. Protons are fundamental constituent

of matter because positive rays are produced by all substances.

“A sub atomic particle, which is a fundamental constituent of all matter

having a mass slightly less than that of H atom and which carries a positive

charge equal in magnitude to the charge on an electron, is called a proton”.

A proton is denoted by p or p+ of +1p.

Comparison of Positive (Canal) Rays and Cathode Rays:

Properties Cathode Rays Canal Rays

Sign of Charge Negative Positive

Mag. of Charge Always –1 Mostly +1, but also +2, +3…

Mass Definite value Variable, depends on ions

e/m Definite value Variable, depends on ions

DISCOVERY OF NEUTRON

After the discovery of electrons and protons. Rutherford (1920) had

predicted the existence of a neutral fundamental particle. In 1932, Chadwick

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bombarded the element Beryllium with particles and noticed the emission

of a radiation having the following characteristics.

The radiation was highly penetrating.

The radiation was unaffected by magnetic and electric fields which

show that it is electrically neutral.

It was found to have approximately the same mass as the protons.

The name „neutron‟ was given to this sub atomic particle. It is denoted by n

or 1

o n . Bombardment of beryllium by particles results in the formation of

carbon and neutrons are emitted.

Be 1

0C n

At present there are a number of evidences which confirm that like electron,

proton and neutron is also a fundamental constituent of atoms (a single

exception is 1

1H atom which does not contain any neutron)

Mass of a neutron is 1.008930 amu (1.6753 10–24g or 1.6753 10–27 kg)

Neutron “A sub atomic particle, which is a fundamental constituent of matter

having mass approximately equal to the hydrogen atom and which is

electrically neutral, is called a neutron”.

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Illustration 1:

The neutron is attracted towards

(A) Positive charged particles

(B) Negative charged particles

(C) Not attracted by any charge

(D) None of these

Solution:

Neutron is an uncharged particle. Hence (C) is correct.

ATOMIC TERM

Nuclide:

Various species of atoms in general.

Nucleons:

Sub-atomic particles in the nucleus of an atom, i.e., protons and

neutrons.

Isotopes:

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Atoms of an element with the same atomic number but different mass

number.

Mass number (A):

Sum of the number of protons and neutrons, i.e., the total number of

nucleons,

Atomic number (Z):

The number of protons in the nucleus of an atom. This, when

subtracted from A, gives the number of neutrons.

Isobars:

Atoms, having the same mass numbers but different atomic numbers,

e.g., 15P32 and 16S

32.

Isotones:

Atoms having the same number of neutrons but different number of

protons or mass number, e.g., 14 16 15

6 8 7C, O, N.

Isoelectronic species:

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Atoms molecules or ions having the same number of electrons, e.g., N2,

CO, CN-.

Nuclear isomers:

Atoms with the same atomic and mass numbers but different

radioactive properties, e.g., uranium X (half life 1.4 min) and uranium Z

(half life 6.7 hours).

Atomic mass unit:

Exactly equal to 1/12th of the mass of 6C12 atom.

(a.m.u.): 1 a.m.u. = 1.66 10–24 g 931.5 MeV

Illustration 2:

The ion that is isoelectronic with CO is

(A) CN- (B) O2+

(C) O2- (D) N2

+

Solution:

Both CO and CN have 14 electrons. Hence (A) is correct

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Illustration 3.

Find the number of neutrons in a neutral atom having atomic mass 23

and number of electrons 11.

Solution:

Number of protons = number of electrons = 11

Number of neutrons = atomic mass – number of protons =23–11 = 12

Illustration 4:

How many protons, electrons and neutrons are present in 0.18 g 30

15 P ?

Solution:

No. of neutrons in one atom = (30 – 15) = 15

0.18 g 30

15 P = 0.18

30 = 0.006 mole

Now, number of 30

15 P atoms in 0.006 mole = 0.006 6.02 1023

Number of electrons in 0.006 mole of 30

15 P =Number of protons in 0.006

mole 30

15 P = 15 0.006 6.02 1023 = 5.418 1022 and number of

neutrons = 5.418 1022

Exercise 1:

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(i) Find the mass number of a uninegative ion having 10 neutrons

and 10 electrons in one mole of the ions.

(ii) Find the total number and total mass of protons present in 34

mg of NH3.

ATOMIC MODELS

We know the fundamental particles of the atom. Now let us see, how

these particles are arranged in an atom to suggest a model of the

atom.

Thomson’s Model:

J.J. Thomson, in 1904, proposed that there was an equal and opposite

positive charge enveloping the electrons in a matrix. This model is called the

plum – pudding model after a type of Victorian dessert in which bits of plums

were surrounded by matrix of pudding.

electron

Positive sphere

This model could not satisfactorily explain the results of scattering

experiment carried out by Rutherford who worked with Thomson.

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Rutherford’s Model:

– particles emitted by radioactive substance were shown to be dipositive

Helium ions (He++) having a mass of 4 units and 2 units of positive charge.

Rutherford allowed a narrow beam of –particles to fall on a very thin gold

foil of thickness of the order of 0.0004 cm and determined the subsequent

path of these particles with the help of a zinc sulphide fluorescent screen.

The zinc sulphide screen gives off a visible flash of light when struck by an

particle, as ZnS has the remarkable property of converting kinetic energy of

particle into visible light. [For this experiment, Rutherford specifically used

particles because they are relatively heavy resulting in high momentum].

Observation:

Majority of the –particles pass straight through the gold strip with

little or no deflection.

Some –particles are deflected from their path and diverge.

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Very few –particles are deflected backwards through angles greater

than 90 .

Some were even scattered in the opposite direction at an angle of 180

[Rutherford was very much surprised by it and remarked that “It was

as incredible as if you fired a 15–inch shell at a piece of tissue paper

and it came back and hit you”]. There is far less difference between air

and bullet than there is between gold atoms and -particle assuming

of course that density of a gold atom is evenly distributed. The

distance of nucleus from where the - particle returns back through

180 is called distance of closet approach and is given by

1 20

2

0

q qr

14 mv

2

Conclusions:

The fact that most of the - particles passed straight through the

metal foil indicates the most part of the atom is empty.

The fact that few - particles are deflected at large angles indicates

the presence of a heavy positively charge body i.e., for such large

deflections to occur - particles must have come closer to or collided

with a massive positively charged body.

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The fact that one in 20,000 have deflected at 180° backwards

indicates that volume occupied by this heavy positively charged body

is very small in comparison to total volume of the atom.

Illustration 5:

An –particle is traveling towards gold nuclei returns back through

10 10 m from it. What is the velocity of the - particle. [Given 1

amu = 1.66 10 27 kg, atomic mass of He = 4 and gold = 79 and

0

1

4 = 9 109 Nm2C 2]

Solution:

We known that 1 2o

2

0

q qr

14 mv

2

2 1 2

0 0

2q qv

4 mr

Now one He atom has charge (q1) = 2e

One gold atom has charge (q2) = 79e

Putting these values we get

19 19 9

2

27 10

2 2 1.6 10 79 1.6 10 9 10v

4 1.66 10 10

v = 3.311 105 m/s

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Conclusions of -Scattering Experiment:

On the basis of the above observation, and having realized that the

rebounding -particles had met something even more massive than

themselves inside the gold atom, Rutherford proposed an atomic model as

follows.

All the +ve charge and nearly the total mass of an atom is present in a

very small region at the centre of the atom. The atom‟s central core is

called nucleus.

The size of the nucleus is very small in comparison to the size of the

atom. Diameter of the nucleus is about 10–13cm while the atom has a

diameter of the order of 10–8 cm. So, the size of atom is 105 times

more than that of nucleus.

Most of the space outside the nucleus is empty.

The electrons, equal in number to the net nuclear positive charge,

revolve around the nucleus with fast speed just like planets around the

sun.

The centrifugal force arising due to the fast speed of an electron

balances the coulombic force of attraction of the nucleus and the

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electron remains stable in its path. Thus according to him atom

consists of two parts (a) nucleus and (b) extra nuclear part.

Defects in Rutherford’s Atomic Model:

Position of electrons: The exact positions of the electrons from the

nucleus are not mentioned.

Stability of the atom: Bohr pointed out that

Rutherford‟s atom should be highly unstable.

According to the law of electro–dynamics, when

a charged body moves under the influence of an

attractive force, it loses energy continuously in

the form of electromagnetic radiation. The

electron should therefore, continuously emit

radiation and lose energy. As a result of this a

moving electron will come closer and closer to

the nucleus and after passing through a spiral

path, it should ultimately fall into the nucleus.

+

It was calculated that the electron should fall into the nucleus in less than

10–8 sec. But it is known that electrons keep moving outside the nucleus.

To solve this problem Neils Bohr proposed an improved form of Rutherford‟s

atomic model.

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Before going into the details of Neils Bohr model we would like to introduce

you some important atomic terms.

Illustration 6:

Prove that density of the nucleus is constant.

Solution:

Radius of the nucleus = 1.33 10–13 A1/3 cm, where A is the mass

number

= 1.33 10–11 A1/3 m

Density of nucleus = Mass

Volume

=27

11 1/ 3 3

A 1.66 10 kg

4(1.33 10 A )

3

= 27

11 3

1.66 10 3

4 3.14 (1.33 10 )kg/m3 = constant

Thus density of nucleus is constant, independent of the element under

consideration.

SOME IMPORTANT CHARACTERISTICS OF A WAVE

A wave is a sort of disturbance which originates from some vibrating source

and travels outward as a continuous sequence of alternating crests and

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troughs. Every wave has five important characteristics, namely, wavelength

( ), frequency ( ), velocity (c), wave number and amplitude (a).

a

Crest Crest

Trough Trough

Electronic Magnetic Radiation:

Ordinary light rays, X–rays, –rays, etc. are called electromagnetic radiations

because similar waves can be produced by moving a charged body in a

magnetic field or a magnet in an electric field. These radiations have wave

characteristics and do not require any medium for their propagation.

Wavelength ( ): The distance between two neighbouring troughs or

crests is known as wavelength. It is denoted by and is expressed in

cm, m, nanometers (1 nm =10–9 m) or Angstrom (1 Å=10–10 m).

Frequency ( ): The frequency of a wave is the number of times a

wave passes through a given point in a medium in one second. It is

denoted by (nu) and is expressed in cycles per second (cps) or hertz

(Hz) 1Hz = 1cps.

The frequency of a wave is inversely proportional to its wave length

( )

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1 or =

c

Velocity: The distance travelled by the wave in one second is called

its velocity. It is denoted by c and is expressed in cm sec–1.

c = or = c

Wave number : It is defined as number of wavelengths per cm. It

is denoted by and is expressed in cm–1.

= 1 or =

c

Amplitude: It is the height of the crest or depth of the trough of a

wave and is denoted by a. It determines the intensity or brightness of

the beam of light.

Electromagnetic Spectrum: The arrangement of the various types of

electromagnetic radiation in order of increasing or decreasing

wavelengths or frequencies is known as electromagnetic spectrum.

Wavelengths of Electromagnetic Radiations:

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Electromagnetic radiations Wave length (Å)

Radio waves 3 1014 to 3 107

Micro waves 3 109 to 3 106

Infrared (IR) 6 106 to 7600

1.1.1 Visible 7600 to 3800

Ultra violet (UV) 3800 to 150

X–rays 150 to 0.1

Gamma rays 0.1 to 0.01

Cosmic rays 0.01 to zero

Illustration 7:

Find out the longest wavelength of absorption line for hydrogen gas

containing atoms in ground state.

Solution:

2

2 2

1 2

1 1 1RZ

n n

For longest wavelength E should be smallest, i.e. transition occurs

from

n = 1 to n = 2

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n = 7

n = 6

n = 5

n = 4

n = 3

n = 2

n = 1

i.e. 1= 109673 cm–1 12

2 2

1 1

1 2

1 = 109673

3

4 cm–1

= 1

4

3 109673cm= 1.2157 10–5 cm = 121.6 nm

Illustration 8:

Calculate the energy in kJ per mole of electronic charge accelerated by

a potential of 1 volt.

Solution:

Energy in joules = charge in coulombs potential difference in volt

= 1.6 10–19 6.02 1023 1

= 9.632 104 J or 96.32 kJ

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Exercise 2:

(i) What is the ratio between energies of two radiations one with a

wavelength of 6000 Å and other with 2000 Å.

(ii) Find the frequency and wave number of a radiation having

wavelength of 1000 Å.

ATOMIC SPECTRUM

If the atom gains energy the electron passes from a lower energy level to a

higher energy level, energy is absorbed that means a specific wave length is

absorbed. Consequently, a dark line will appear in the spectrum. This dark

line constitutes the absorption spectrum.

If the atom loses energy, the electron passes from higher to a lower energy

level, energy is released and a spectral line of specific wavelength is

emitted. This line constitutes the emission spectrum.

Types of Emission Spectra:

Continuous spectra: When white light from any source such as sun

or bulb is analyzed by passing through a prism, it splits up into seven

different wide bands of colour from violet to red (like rainbow). These

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colour are so continuous that each of them merges into the next.

Hence the spectrum is called as continuous spectrum.

Line spectra: When an electric discharge is passed through a gas at

low pressure light is emitted. If this light is resolved by a

spectroscope, It is found that some isolated coloured lines are

obtained on a photographic plate separated from each other by dark

spaces. This spectrum is called line spectrum. Each line in the

spectrum corresponds to a particular wavelength. Each element gives

its own characteristic spectrum.

PLANCK’S QUANTUM THEORY

When a black body is heated, it emits thermal radiations of different

wavelengths or frequency. To explain these radiations, Max Planck put

forward a theory known as Planck‟s quantum theory. The main points of

quantum theory are

Substances radiate or absorb energy discontinuously in the form of

small packets or bundles of energy.

The smallest packet of energy is called quantum. In case of light the

quantum is known as photon.

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The energy of a quantum is directly proportional to the frequency of

the radiation. E (or) E = h where is the frequency of radiation

and h is Planck‟s constant having the value 6.626 10–27 erg – sec or

6.626 10–34 J–sec.

A body can radiate or absorb energy in whole number multiples of a

quantum h , 2h ,3h ………..nh . where „n' is the positive integer.

Neils Bohr used this theory to explain the structure of atom.

Illustration 9.

The wave number of a radiation is 400 cm 1. Find out its

(a) Wavelength (b) Frequency

(c) J per photon (d) kcal per mol of photons

(e) kJ per mol of photons

Solution:

(a) 1or =

1 =

1

1

400 cm = 2.5 10–3 cm

(b) = c = c = 3 10–10 cm/s 400 cm–1 = 7 11.2 10 s

(c) Ephoton = h = hc

= h c

= 6.626 10–34 Js 3 108 cm/s 400 cm–1

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= 7.95 10–21J

(d) Ephoton = 7.95 10–21J

For 1 mol of photon energy = 7.95 10–21J 6.022 1023 mol–1

= (4.7875 103 J mol–1) (1 kcal/4184 J)

= 1.14 kcal mol–1

(e) E = (4.7875 103 J mol–1) (1 kJ/1000J)

= 4.7875 kJ mol–1

Illustration 10:

A near ultraviolet photon of 300 nm is absorbed by a gas and then

re emitted as two photons. One photon is red with wavelength 760

nm. What would be the wavelength of the second photon?

Solution:

E = hc

Wavelength of the 1st photon = 1

E1 = 1

hc

And that of second photon = 2

E2 = 2

hc

ETotal = E1 + E2= emitted energy

1 2

hc hc hc

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2

1 1 1

300 760

2

1 1 1

300 760

2

1 760 300

760 300

2 = wavelength of the second photon = 496 nm

BOHR’S ATOMIC MODEL

Bohr developed a model for hydrogen atom and hydrogen like one–electron

species (hydrogenic species). He applied quantum theory in considering the

energy of an electron bound to the nucleus.

Important Postulates:

An atom consists of a dense nucleus situated at the centre with the

electron revolving around it in circular orbits without emitting any

energy. The force of attraction between the nucleus and an electron is

equal to the centrifugal force of the moving electron.

Of the finite number of circular orbits around the nucleus, an electron

can revolve only in those orbits whose angular momentum (mvr) is an

integral multiple of factor h 2

mvr = nh

2

where, m = mass of the electron

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v = velocity of the electron

n = orbit number in which electron is present

r = radius of the orbit

As long as an electron is revolving in an orbit it neither loses nor gains

energy. Hence these orbits are called stationary states. Each

stationary state is associated with a definite amount of energy and it is

also known as energy levels. The greater the distance of the energy

level from the nucleus, the more is the energy associated with it. The

different energy levels are numbered as 1,2,3,4, (from nucleus

onwards) or K, L, M, N etc.

Ordinarily an electron continues to move in a particular stationary

state without losing energy. Such a stable state of the atom is called

as ground state or normal state.

If energy is supplied to an electron, it may jump (excite)

instantaneously from lower energy (say 1) to higher energy level (say

2, 3, 4, etc.) by absorbing one or more quanta of energy. This new

state of electron is called as excited state. The quantum of energy

absorbed is equal to the difference in energies of the two concerned

levels.

Since the excited state is less stable, atom will lose it‟s energy and

come back to the ground state.

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Energy absorbed or released in an electron jump, ( E) is given by

E = E2 – E1 = h

where E1 and E2 are the energies of the electron in the first and second

energy levels, and is the frequency of radiation absorbed or emitted.

Note:

If the energy supplied to hydrogen atom is less than 13.6 eV, it will

accept or absorb only those quanta which can take it to a certain

higher energy level i.e., all those photons having energy less than or

more than a particular energy level will not be absorbed by hydrogen

atom. But if energy supplied to hydrogen atom is more than 13.6 eV

then all photons are absorbed and excess energy appears as kinetic

energy of emitted photo electron.

Radius and Energy Levels of Hydrogen Atom:

Consider an electron of mass „m‟ and charge „e‟ revolving around a nucleus

of charge Ze (where, Z = atomic number and e is the charge of the proton)

with a tangential velocity v. r is the radius of the orbit in which electron is

revolving.

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By Coulomb‟s Law, the electrostatic force of attraction between the moving

electron and nucleus is Coulombic force = 2

2

KZe

r

K = o

1

4 (where o is permittivity of free space)

K = 9 109 Nm2 C–2

In C.G.S. units, value of K = 1 dyne cm2 (esu)–2

The centrifugal force acting on the electron is 2mv

r

Since the electrostatic force balances the centrifugal force, for the stable

electron orbit.

2mv

r=

2

2

KZe

r … (1)

or v2 = 2KZe

mr … (2)

According to Bohr‟s postulate of angular momentum quantization, we have

mvr =nh

2

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v = nh

2 mr

v2 =2 2

2 2 2

n h

4 m r … (3)

Equating (2) and (3)

2 2 2

2 2 2

KZe n h

mr 4 m r

Solving for r we get r = 2 2

2 2

n h

4π mΚΖe

Where n = 1, 2, 3 - - - - -

Hence only certain orbits whose radii are given by the above equation are

available for the electron. The greater the value of n, i.e., farther the energy

level from the nucleus the greater is the radius.

The radius of the smallest orbit (n=1) for hydrogen atom (Z=1) is ro.

ro =2 2

2 2

n h

4 me K=

22 34

22 31 19 9

1 6.626 10

4 3.14 9 10 1.6 10 9 10=5.29 10–11 m=0.529 Å

Radius of nth orbit for an atom with atomic number Z is simply written as

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rn = 0.529 2n

ZÅ

Calculation of Energy of an Electron:

The total energy, E of the electron is the sum of kinetic energy and potential

energy.

Kinetic energy of the electron = ½ mv2

Potential energy = 2KZe

r

Total energy = 1/2 mv2 –2KZe

r … (4)

From equation (1) we know that

2mv

r=

2

2

KZe

r

½ mv2 = 2KZe

2r

Substituting this in equation (4)

Total energy (E) =2KZe

2r–

2KZe

r =

2KZe

2r

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Substituting for r, gives us

E =2 2 4 2

2 2

2π mZ e K

n h where n = 1, 2, 3……….

This expression shows that only certain energies are allowed to the electron.

Since this energy expression consist of so many fundamental constant, we

are giving you the following simplified expressions.

E = –21.8 10–12 2

2

Z

n erg per atom

= –21.8 10–19 2

2

Z

n J per atom

= –13.6 2

2

Z

n eV per atom

(1eV = 3.83 10–23 kcal

1eV = 1.602 10–12 erg

1eV = 1.602 10–19J)

E = –313.6 2

2

Z

nkcal / mole (1 cal = 4.18 J)

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The energies are negative since the energy of the electron in the atom is

less than the energy of a free electron (i.e., the electron is at infinite

distance from the nucleus) which is taken as zero. The lowest energy level of

the atom corresponds to n=1, and as the quantum number increases, E

becomes less negative.

When n = , E = 0, which corresponds to an ionized atom i.e., the electron

and nucleus are infinitely separated.

H H++ e– (ionization).

Calculation of Velocity:

We know that

mvr =nh

2; v =

nh

2 mr

By substituting for r we get

v = 22 KZe

nh

Where except n and Z all are constants

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v = 2.18 108 Z

ncm/sec.

Further application of Bohr‟s work was made, to other one electron species

(Hydrogenic ion) such as He+ and Li2+. In each case of this kind, Bohr‟s

prediction of the spectrum was correct.

Illustration 11:

The velocity of electron in the second orbit of He will be

(A) 62.18 10 m/ s (B) 61.09 10 m/ s

(C) 64.36 10 m/ s (D) None of these

Solution:

0n

v Zv

n. Hence (A) is correct.

Illustration 12:

Calculate the velocity of an electron in Bohr‟s first orbit of hydrogen

atom

(Given r = 0.53 10–10 m)

Solution:

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According to Bohr‟s theory

mvr = nh nh

or v2 2 mr

r = 0.53 10–10m

n = 1, h = 6.62 10–34kg m2 s–1, m = 9.1 10–31 kg

v = 34 2 1

31 10

1 (6.62 10 kgm s )

222 (9.1 10 kg) (0.53 10 m)

7

= 2.18 106 m/s

Illustration 13:

The velocity of e in first Bohr‟s orbit is 2.17 106 m/s. Calculate the

velocity in 3rd orbit of He2+ ion.

Solution:

vn = 0

Zv

n

Hence v = 62.19 10 2

3 = 1.45 106 m/s

HYDROGEN ATOM

If an electric discharge is passed through hydrogen gas taken in a discharge

tube under low pressure, and the emitted radiation is analyzed with the help

of spectrograph, it is found to consist of a series of sharp lines in the UV,

visible and IR regions. This series of lines is known as line or atomic

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spectrum of hydrogen. The lines in the visible region can be directly seen on

the photographic film.

Each line of the spectrum corresponds to a light of definite wavelength. The

entire spectrum consists of six series of lines, each series, known after their

discoverer as the Balmer, Paschen, Lyman, Brackett, Pfund and Humphrey

series. The wavelength of all these series can be expressed by a single

formula.

1= R

2 2

1 2

1 1

n n

Where, = wave number

= wave length

R = Rydberg constant (109678 cm–1)

n1 and n2 have integral values as follows

Series n1 n2 Main spectral lines

Lyman

Balmer

Paschen

Brackett

Pfund

1

2

3

4

5

2, 3, 4, etc

3, 4, 5 etc

4, 5, 6 etc

5, 6, 7 etc

6, 7, etc

Ultra – violet

Visible

Infra – red

Infra – red

Infra – red

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Note:

All lines in the visible region are of Balmer series but reverse is not

true. i.e., all Balmer lines will not fall in visible region]

The pattern of lines in atomic spectrum is characteristic of hydrogen.

Illustration 14:

Find the wavelength of a spectral line produced when an electron in

the H-atoms jumps from 4th level to 2nd level.

Solution:

H 2 2

1 2

1 1 1R

n n

1

2 2

1 1 1109678 cm

2 4

16cm

3 109678

= 4863 0

A

Merits of Bohr’s Theory:

The experimental value of radii and energies in hydrogen atom are in

good agreement with that calculated on the basis of Bohr‟s theory.

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Bohr‟s concept of stationary state of electron explains the emission

and absorption spectra of hydrogen like atoms.

The experimental values of the spectral lines of the hydrogen spectrum

are in close agreement with that calculated by Bohr‟s theory.

Limitations of Bohr’s Theory

It does not explain the spectra of atoms having more than one

electron.

Bohr‟s atomic model failed to account for the effect of magnetic field

(Zeeman effect) or electric field (Stark effect) on the spectra of atoms

or ions. It was observed that when the source of a spectrum is placed

in a strong magnetic or electric field, each spectral line further splits

into a number of lines. This observation could not be explained on the

basis of Bohr‟s model.

De Broglie suggested that electrons like light have dual character. It

has particle and wave character. Bohr treated the electron only as

particle.

Another objection to Bohr‟s theory came from Heisenberg‟s

Uncertainty Principle. According to this principle “It is impossible to

determine simultaneously the exact position and momentum of a small

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moving particle like an electron”. The postulate of Bohr, that electrons

revolve in well defined orbits around the nucleus with well defined

velocities is thus not tenable.

Illustration 15:

The radii of two of the first four Bohr orbits of the hydrogen atom are

in the ratio 1 : 4. The energy difference between them may be

(A) 0.85 eV (B) 2.55 eV

(C) 3.40 eV (D) 8.20 eV

Solution:

2

0n

r nr

Zand

2

0n 2

E ZE

n. Hence (B) is correct.

Illustration 16:

An electron in a hydrogen atom in its ground state absorbs 1.50 times

as much energy as the minimum required for its escape from the

atom. What is the wavelength of the emitted electron? (me = 9.11

10–31 kg)

(A) 4.7 Å (B) 4.7 nm

(C) 9.4 Å (D) 9.40 nm

Solution:

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Since 13.6 eV is need for ionization

Total energy absorbed = 1.5 13.6eV 20.4eV

6.8 eV is converted into K.E.

34

31 19

h 6.625 10

2m KE 2 9.11 10 6.8 1.6 10= 104.7 10 m 4.7Å

Hence (A) is correct.

Illustration 17:

The observation that electrons can be diffracted is an evidence that

electrons

(A) have particle properties (B) have wave properties

(C) are emitted by atoms (D) are absorbed by ions

Solution:

Diffraction prove wave character of electron. Hence (B) is correct.

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Illustration 18:

A series of lines in the spectrum of atomic hydrogen lies at

wavelengths 656.46, 482.7, 434.17, 410.29 nm. What is the

wavelength of next line in this series?

Solution:

The given series of lines are in the visible region and thus appears to

be Balmer series

Therefore n1 = 2 and n2=? for next line

If = 410.29 10–7 cm and n1 = 2

n2 may be calculated for the last line

1= R

2 2

1 2

1 1

n n

7

1

410.29 10 = 109673

2 2

2

1 1

2 n

n2 = 6

Thus next line will be obtained during the jump of electron from 7th to

2nd shell i.e,

1= R

2 2

1 1

2 7 = 109673

1 1

4 49

= 397.2 10–7 cm = 397.2 nm

Illustration 19:

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Find the energy released in (ergs) when 2.0 gm atom of hydrogen

undergo transition giving spectral line of lowest energy in visible

region of its atomic spectrum.

Solution:

E = 2.178 10–18 (Z2) 2 2

1 2

1 1J/ atom

n n

For visible photon, n1 = 2

For lowest energy transition n2 = 3

E = 2.178 10–18 12 2 2

1 1

2 3J/atom

For 2.0g atom

E = (2 6.023 1023) 2.178 10–18 5

36joules

= 3.63 105J = 3.63 1012 ergs

Illustration 20:

Which hydrogen like ionic species has wavelength difference between

the first line of Balmer and first line of Lyman series equal to 59.3

10–9 m? Neglect the reduced mass effect.

Solution:

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Wave number of first Balmer line of an species with atomic number Z

is given by

2

2 2

1 1RZ

2 3

25RZ

36; Similarly wave number of v of first Lyman line is given by

= RZ22 2

1 1

1 2 = 23RZ

4;

1 1and

– = 2 2

36 4

5RZ 3RZ =

2

1 36 4

5 3RZ =

2

88

15RZ

Z2 = 9 7

88

59.3 10 15 1.097 10 = 9

or Z = 3

ionic species is Li2+

Exercise 3:

(i) Calculate of the radiations when the electron jumps from III to II

orbit of hydrogen atom. The electronic energy in II and III Bohr orbit

of hydrogen atoms are –5.42 10–12 and –2.41 10–12 erg

respectively.

(ii) In a hydrogen atom, the energy of an electron in first Bohr‟s orbit is –

13.12 105 J mol–1. What is the energy required for its excitation to

Bohr‟s second orbit.

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(iii) What is the wavelength of the light emitted when the electron in a

hydrogen atom undergoes transition from an energy level with n = 4

to an energy level with n =? What is the colour corresponding to this

wavelength? (Rydberg constant = 109,677 cm 1)

QUANTUM NUMBERS

An atom contains large number of shells and subshells. These are

distinguished from one another on the basis of their size, shape and

orientation (direction) in space. The parameters are expressed in terms of

different numbers called quantum numbers.

Quantum numbers may be defined as a set of four numbers with the help of

which we can get complete information about all the electrons in an atom. It

tells us the address of the electron i.e., location, energy, the type of orbital

occupied and orientation of that orbital.

Principal quantum number (n): It tells the main shell in which the

electron resides, the approximate distance of the electron from the

nucleus and energy of that particular electron. It also tells the

maximum number of electrons that a shell can accommodate is 2n2,

where n is the principal quantum number.

Shell K L M N

Principal quantum number (n) 1 2 3 4

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Maximum number of electrons 2 8 18 32

Azimuthal or angular momentum quantum number : This

represents the number of subshells present in the main shell. These

subsidiary orbits within a shell will be denoted as 0, 1, 2, 3, 4,… or s,

p, d, f… This tells the shape of the subshells. The orbital angular

momentum of the electron is given as 1 h2

or 1 for a

particular value of „n‟h

where 2

. For a given value of n values of

possible vary from 0 to n – 1.

The magnetic quantum number (m): An electron due to its angular

motion around the nucleus generates an electric field. This electric

field is expected to produce a magnetic field. Under the influence of

external magnetic field, the electrons of a subshell can orient

themselves in certain preferred regions of space around the nucleus

called orbitals. The magnetic quantum number determines the number

of preferred orientations of the electron present in a subshell. The

values allowed depends on the value of l, the angular momentum

quantum number, m can assume all integral values between – to +

including zero. Thus m can be –1, 0, +1 for = 1. Total values of m

associated with a particular value of is given by 2 + 1.

The spin quantum number (s): Just like earth which not only

revolves around the sun but also spins about its own axis, an electron

in an atom not only revolves around the nucleus but also spins about

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its own axis. Since an electron can spin either in clockwise direction or

in anticlockwise direction, therefore, for any particular value of

magnetic quantum number, spin quantum number can have two

values, i.e., +1/2 and –1/2 or these are represented by two arrows

pointing in the opposite directions, i.e., and . When an electron

goes to a vacant orbital, it can have a clockwise or anti clockwise spin

i.e., +1/2 or –1/2. This quantum number helps to explain the

magnetic properties of the substances.

Illustration 21:

If the principal quantum number n has a value of 3, what are

permitted values of the quantum numbers „‟ and „m‟?

Solution:

If principal quantum number (n) has the value of 3, subsidary

quantum number () will also have three values i.e. 0, 1 and 2 and

magnetic quantum number (m) will have n2, i.e. 32 i.e. 9 values in all

and they are designated as under:

n = 3 = 0 m = 0 (s)

= 1 m = – 1 (p)

m = 0

m = + 1

= 2 m = – 2 (d)

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m = – 1

m = 0

m = +1

m = +2

Illustration 22:

In which orbital the electron will reside if it has the following values of

quantum numbers.

(i) n = 3, 1, m 0

(ii) n = 4, 0, m 0

Solution:

(i) 1 for p subshell and m = 0 for pz orbital.

Hence electron will residue in 3pz orbital.

(ii) 0 for s subshell. Hence electron will residue in 4s orbital.

Illustration 23:

Write all the quantum numbers for the following orbitals.

(i) 3 2zd (ii) 3s (iii) 4px

Solution:

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(i) n 3, 2, m 0, s 1/2

(ii) n 3, 0, m 0, s 1/2

(iii) n 4, 1, m 1or 1, s 1/2

Pauli’s Exclusion Principle:

According to this principle, an orbital can contain a maximum number of two

electrons and these two electrons must be of opposite spin.

Two electrons in an orbital can be represented by

or

SHAPES AND SIZE OF ORBITALS

An orbital is the region of space around the nucleus within which the

probability of finding an electron of given energy is maximum (90–95%).

The shape of this region (electron cloud) gives the shape of the orbital. It is

basically determined by the azimuthal quantum number , while the

orientation of orbital depends on the magnetic quantum number (m). Let us

now see the shapes of orbitals in the various subshells.

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s–orbitals: These orbitals are spherical and

symmetrical about the nucleus. The

probability of finding the electron is

maximum near the nucleus and keep on

decreasing as the distance from the nucleus

increases. There is vacant space between

two successive s–orbitals known as radial

node. But there is no radial node for 1s

orbital since it is starting from the nucleus.

nucleus

Z radial node

2s 1s

x

y

The size of the orbital depends upon the value of principal quantum number

(n). Greater the value of n, larger is the size of the orbital. Therefore,

2s–orbital is larger than 1s orbital but both of them are non-directional and

spherically symmetrical in shape.

p–orbitals ( =1):

The probability of finding the p–electron is maximum in two lobes on the

opposite sides of the nucleus. This gives rise to a dumb–bell shape for the

p–orbital. For p–orbital = 1. Hence, m = –1, 0, +1. Thus, p–orbital have

three different orientations. These are designated as px, py & pz depending

upon whether the density of electron is maximum along the x y and z axis

respectively. As they are not spherically symmetrical, they have directional

character. The two lobes of p–orbitals are separated by a nodal plane, where

the probability of finding electron is zero.

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Y

pz py

px Z

X

The three p-orbitals belonging to a particular energy shell have equal

energies and are called degenerate orbitals.

d–orbitals ( = 2):

For d–orbitals, l = 2. Hence m = –2,–1, 0, +1, +2. Thus there are 5d

orbitals. They have relatively complex geometry. Out of the five orbitals, the

three (dxy, dyz,dzx) project in between the axis and the other two 2zd and

2 2yd

x lie along the axis.

Y

X Z

2 2x yd

2zd

Dough–nut shape or Baby soother shape

Clover leaf shape

Z Y

dxy

X X Y

Z

dyz dxz

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Illustration 24:

An electron has a spin quantum number 1

2and a magnetic quantum

number 1. It cannot occupy a/an

(A) d orbital (B) f orbital

(C) p orbital (D) s orbital

Solution:

For s orbital m = 0. Hence (D) is correct.

Illustration 25:

Which d-orbital has electron density in all the planes?

Solution:

2zd orbital has lobe along Z-axis and a belt of electrons in xy plane.

Hence it has electron density in all xy, xz and yz plane.

RULES FOR FILLING OF ELECTRONS IN VARIOUS ORBITALS

The atom is built up by filling electrons in various orbitals according to the

following rules.

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Aufbau Principle: This principle states that the electrons are added one by

one to the various orbitals in order of their increasing energy starting with

the orbital of lowest energy. The increasing order of energy of various orbital

is 1s,2s,2p,3s,3p,4s,3d,4p,5s,4d,5p,6s,4f,5d,6p,5f,6d,7p……………………

How to remember such a big sequence? To make it simple we are giving

you the method to write the increasing order of the orbitals. Starting from

the top, the direction of the arrow gives the order of filling of orbitals.

1s

2s

3s

4s

5s

6s

7s

2p

3p

4p

5p

6p

3d

4d

5d

4f

Alternatively, the order of increasing energies of the various orbitals can be

calculated on the basis of (n + ) rule.

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The energy of an orbital depends upon the sum of values of the principal

quantum number (n) and the azimuthal quantum number ( ). This is called

(n + ) rule. According to this rule,

“In neutral isolated atom, the lower the value of (n + ) for an orbital, lower

is its energy. However, if the two different types of orbitals have the same

value of (n + ), the orbitals with lower value of n has lower energy‟‟.

Illustration of (n + ) Rule:

Type of

orbitals

Value

of n

Values

of

Values of

(n+ ) Relative energy

1s 1 0 1 + 0 = 1 Lowest energy

2s 2 0 2 + 0 = 2 Higher energy than 1s orbital

2p 2 1 2 + 1 = 3 2p orbital (n=2) have lower energy

than 3s orbital (n=3) 3s 3 0 3 + 1 = 4

Hund’s Rule of Maximum Multiplicity:

This rule deals with the filling of electrons in the equal energy (degenerate)

orbitals of the same sub shell (p, d and f). According to this rule,

“Electron pairing in p,d and f orbitals cannot occur until each orbital of a

given subshell contains one electron each or is singly occupied”.

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This is due to the fact that electrons being identical in charge, repel each

other when present in the same orbital. This repulsion can, however, be

minimized if two electrons move as far apart as possible by occupying

different degenerate orbitals. All the electrons in a degenerate set of orbitals

will have same spin.

Electronic Configuration of Elements:

Electronic configuration is the distribution of electrons into different shells,

subshells and orbitals of an atom.

Keeping in view the above mentioned rules, electronic configuration of any

orbital can be simply represented by the notation]

n x

Number of electrons in the subshell

Symbol of subshell or orbitals (s,p,d,f)

Principal quantum number

Alternatively:

Orbital can be represented by a box and an electron with its direction of spin

by arrow. To write the electronic configuration, just we need to know (i) the

atomic number (ii) the order in which orbitals are to be filled (iii) maximum

number of electrons in a shell, sub–shell or orbital.

Each orbital can accommodate two electrons

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The number of electrons to be accomodated in a subshell is 2

numbers of degenerate orbitals.

Subshell Maximum number of electrons

s 2

p 6

d 10

f 14

The maximum number of electrons in each shell (K, L, M, N…) is given

by 2n2 where n is the principal quantum number.

The maximum number of orbitals in a shell is given by n2 where n is

the principal quantum number.

Importance of Knowing the Electronic Configuration:

The chemical properties of an element are dependent on the relative

arrangement of its electrons.

Illustration 26:

Write the electronic configuration of nitrogen (atomic number = 7)

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Solution:

1s2 2s

2 2p

3

Illustration 27:

Write the electronic configuration of following:

(i) 2 2S z 16 (ii) Fe z 26

Solution:

(i) 2 2 6 2 61s 2s 2p 3s 3p

(ii) 2 2 6 2 6 61s 2s 2p 3s 3p 3d

Exceptional Configurations:

Stability of Half Filled and Completely Filled Orbitals:

Cu has 29 electrons. Its expected electronic configuration is

1s22s22p63s23p64s23d9

But a shift of one electron from lower energy 4s orbital to higher energy 3d

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orbital will make the distribution of electron symmetrical and hence will

impart more stability.

Thus the electronic configuration of Cu is

1s2 2s22p63s23p64s13d10

Fully filled and half filled orbitals are more stable.

Illustration 28:

Which of the following metals has the highest value of exchange

energy required for exchange stabilization?

(A) Mn (B) Cr

(C) Cu (D) Zn

Solution:

Cu has full filled d-subshell. Hence (C) is correct.

Illustration 29:

The total spin resulting from a d7 configuration is

(A) 1 (B) 2

(C) 52 (D) 3

2

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Solution:

S = n/2, where n = number of unpaired electrons = 3. Hence (D) is

correct.

Illustration 30:

The electronic configuration of an element is 1s22s22p63s23p63d54s1.

This represents its

(A) excited state (B) ground state

(C) cationic form (D) anionic form

Solution:

It is ground state electronic configuration of Cr. Hence (B) is correct.

Illustration 31:

Write the electronic configuration of the following ions:

(A) H+ (B) Na+

(C) O2 (D) F

Solution:

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(A) 1s0

(B) 1s22s22p6

(C) 1s22s22p6

(D) 1s22s22p6

Illustration 32:

We know the Hund‟s rule. Explain how to arrange three electrons in p

orbitals.

Solution:

The important point to be remembered is, all the singly occupied

orbitals should have parallel spins, i.e in the same direction either-

clockwise or anticlockwise.

2px 2py 2pz

½ ½ ½ = 1½

The maximum multiplicity means that the total spin of unpaired

electrons is maximum.

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Illustration 33:

We know that fully filled and half filled orbitals are more stable. Can

you write the electronic configuration of Cr(Z = 24)?.

Solution:

Cr (Z = 24)

1s2, 2s2,2p6,3s2,3p6,4s1,3d5.

Since half filled orbital is more stable one 4s electron is shifted to 3d

orbital.

Illustration 34:

The configuration of potassium (19) is 1s22s22p63s24s1 and not

1s22s22p63s23p63d1. Why?

Solution:

According to Aufbau Principle‟s electron first fill in a sub-level of lower

energy level (lower n+). The energy of 4s(4 + 0 = 4) is lower than

that of 3d (3 + 2 = 5), so the electronic configuration of K is

1s22s22p63s23p64s1.

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Illustration 35:

How many maximum number of electrons can be present in 3rd shell of

an atom?

Solution:

Total number of electrons in a shell = 2n2

= 2 32 = 18

Illustration 36:

How many maximum number of electrons a d-orbital can have?

Solution:

Any orbital can have maximum of 2 electrons.

Illustration 37:

Calculate the total number of d-electrons in molybdenum (atomic

number = 42).

Solution:

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Electronic configuration of Mo is

2 6 2 6 2 10 6 2 41s 2p 3s 3p 4s 3d 4p 5s 4d

Hence, total number of d-electrons = 10 + 4 = 14

Exercise 4:

(i) Write the electronic configuration of the following:

(a) Cr3+

(b) Mn2+

(c) Cu

(ii) Write the orbital rotation for the following quantum numbers.

(a) n = 4, = 2

(b) n = 2, = 1

(c) n = 3, = 0

DUAL CHARACTER (PARTICLE AND WAVE CHARACTER OF MATTER

AND RADIATION)

In case of light some phenomenon like diffraction and interference can be

explained on the basis of its wave character. However, the certain other

phenomenon such as black body radiation and photoelectric effect can be

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explained only on the basis of its particle nature. Thus, light is said to have a

dual character. Such studies on light were made by Einstein in 1905.

Louis de Broglie, in 1924 extended the idea of photons to material particles

such as electron and he proposed that matter also has a dual character-as

wave and as particle.

Derivation of de-Broglie Equation:

The wavelength of the wave associated with any material particle was

calculated by analogy with photon.

In case of photon, if it is assumed to have wave character, its energy is

given by

E = h …(i)

(according to the Planck‟s quantum theory)

where is the frequency of the wave and „h‟ is Planck‟s constant

If the photon is supposed to have particle character, its energy is given by

E = mc2 … (ii)

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(according to Einstein‟s equation)

where „m‟ is the mass of photon, „c‟ is the velocity of light.

By equating (i) and (ii)

h = mc2

But = c/

h c = mc2

(or) = h /mc

The above equation is applicable to material particle if the mass and velocity

of photon is replaced by the mass and velocity of material particle. Thus for

any material particle like electron.

= h/mv or = h

p where mv = p is the momentum of the particle.

Relation Between Kinetic Energy and Wavelength:

K.E (E)=1/2 mv2

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v = 2E

m

h

mv

h

2E.m

Derivation of Angular Momentum from de Broglie Equation:

According to Bohr‟s model, the electron revolves around the nucleus in

circular orbits. According to de Broglie concept, the electron is not only a

particle but has a wave character also.

If the wave is completely in phase, the circumference of

the orbit must be equal to an integral multiple of wave

length ( )

Therefore 2 r = n

where „n‟ is an integer and „r‟ is the radius of the orbit

But = h/mv

2 r = nh /mv or mvr = n h/2

which is Bohr‟s postulate of angular momentum, where „n‟ is the principal

quantum number.

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“Thus, the number of waves an electron makes in a particular Bohr orbit in

one complete revolution is equal to the principal quantum number of the

orbit”.

Alternatively:

Number of waves „n‟ = 2 r

= 2 r

h mv =

2 mvr

h

Where v and r are the velocity of electron and radius of that particular Bohr

orbit in which number of waves are to be calculated, respectively.

The electron is revolving around the nucleus in a circular orbit. How many

revolutions it can make in one second?

Let the velocity of electron be v m/sec. The distance it has to travel for one

revolution 2 r, (i.e., the circumference of the circle).

Thus, the number of revolutions per second is = v

2 r

Common unit of energy is electron volt which is amount of energy given

when an electron is accelerated by a potential of exactly 1 volt. This energy

equals the product of voltage and charge. Since in SI units

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coulombs volts = joules, 1 eV numerically equals the electronic charge

except that joule replaces coulombs.

Illustration 38:

An e-, a proton and an alpha particle have K.E of 16 E, 4 E and E

respectively. What‟s the qualitative order of their Broglie wavelengths?

(A) e > P > a (B) e > p = a

(C) P < e < a (D) a < e = P

Solution:

=h

2mK.E. Hence (B) is correct.

Illustration 39:

Calculate the de Broglie wavelength of a ball of mass 0.1 kg moving

with a speed of 60 ms–1.

Solution:

h

mv=

346.6 10

0.1 60

= 1.1 10–34 m.

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This is apparent that this wavelength is too small for ordinary observation.

Although the de Broglie equation is applicable to all material objects but it

has significance only in case of microscopic particles.

Since, we come across macroscopic objects in our everyday life, de Broglie

relationship has no significance in everyday life.

[Distinction between the wave- particle nature of a photon and the

particle- wave nature of a sub atomic particle]:

Photon Sub Atomic Particle

1. Energy = h Energy = 1

2mv2

2. Wavelength = c Wavelength =

h

mv

Note:

We should never interchange any of the above

Illustration 40:

Calculate the number of waves made by electron in 4th Bohr‟s orbit.

Solution:

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Number of waves made by electron in nth orbit = n

Hence the number of waves in 4rd orbit = 4

Illustration 41:

What is the de-Broglie wavelength of electron having K.E. of 5 eV?

Solution:

K.E. = 21mv

2

2K.E.v

m

hNow,

mv

= h

2K.E. m

= 34

19 31

6.62 10

2 5 1.6 10 9.1 10

= 5.486 10 10 m

Illustration 42:

Through what potential difference an electron must be accelerated to

have a de-Broglie wavelength of 1o

A .

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Solution:

= 10 10 m

We know K.E. = 21mv eV

2

v = 2eV

m

Now h h

mv 2eVm

Hence 2

2

hv

2 em

234

10 2 19 31

6.62 10

2 10 h 1.6 10 9.1 10= 1.5 106 volt

HEISENBERG’S UNCERTAINTY PRINCIPLE

All moving objects that we see around us e.g., a car, a ball thrown in the air

etc., move along definite paths. Hence their position and velocity can be

measured accurately at any instant of time. Is it possible for subatomic

particle also?

As a consequence of dual nature of matter, Heisenberg, in 1927 gave a

principle about the uncertainties in simultaneous measurement of position

and momentum (mass velocity) of small particles.

This Principle States:

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“It is impossible to measure simultaneously the position and momentum of a

small microscopic moving particle with absolute accuracy or certainty” i.e., if

an attempt is made to measure any one of these two quantities with higher

accuracy, the other becomes less accurate.

The product of the uncertainty in position ( x) and the uncertainty in the

momentum ( p = m. v where m is the mass of the particle and v is the

uncertainty in velocity) is equal to or greater than h/4 where h is the

Planck‟s constant.

Thus, the mathematical expression for the Heisenberg‟s uncertainty principle

is simply written as

x. p h/4

Explanation of Heisenberg’s uncertainty principle

Suppose we attempt to measure both the position and momentum of an

electron, to pinpoint the position of the electron we have to use light so that

the photon of light strikes the electron and the reflected photon is seen in

the microscope. As a result of the hitting, the position as well as the velocity

of the electron are disturbed. The accuracy with which the position of the

particle can be measured depends upon the wavelength of the light used.

The uncertainty in position is . The shorter the wavelength, the greater is

the accuracy. But shorter wavelength means higher frequency and hence

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higher energy. This high energy photon on striking the electron changes its

speed as well as direction. But this is not true for macroscopic moving

particle. Hence Heisenberg‟s uncertainty principle is not applicable to

macroscopic particles.

Illustration 43:

Why electron cannot exist inside the nucleus according to Heisenberg‟s

uncertainty principle?

Solution:

Diameter of the atomic nucleus is of the order of 10–15m

The maximum uncertainty in the position of electron is 10–15 m.

Mass of electron = 9.1 10–31 kg.

x. p = h

4

x (m. v) = h/4

v = h 1

4 x.m =

346.63 10

224

7

15 31

1

10 9.1 10

v = 5.80 1010 ms–1

This value is much higher than the velocity of light and hence not

possible.

Illustration 44:

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What is the uncertainity in the position of electron, if uncertainity in its

velocity is 0.0058 m/s?

Solution:

hx v

4 m

34

31

6.02 10x

4 3.14 9.1 10 0.0058

x = 0.01 m

Illustration 45:

What is the uncertainity in the position of a ball of mass 10 g and

which is moving with a velocity of 100 m/s with 0.002 % uncertainity?

Solution:

hx v

4 m

Now 100 0.002

v 0.002 m/s100

34

31

6.02 10x 0.0289 m

4 3.14 9.1 10 0.002

Exercise 5:

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(i) Two particles A and B are in motion. If the wavelength associated

with particle A is 5 10–8 m, calculate the wavelength associated with

particle B if its momentum is half of A.

(ii) Calculate the de Broglie wavelength of an electron that has been

accelerated from rest through a potential difference of 1 kV.

(iii) Calculate the uncertainty in position of a particle when uncertainty in

the momentum is

(a) 1 10–2 gm cm sec–1 and (b) zero.

QUANTUM MECHANICAL MODEL OF ATOM

The atomic model which is based on the particle and wave nature of the

electron is known as wave or quantum mechanical model of the atom. This

was developed by Erwin Schrodinger in 1926. This model describes the

electron as a three dimensional wave in the electronic field of positively

charged nucleus. Schrodinger derived an equation which describes wave

motion of an electron. The differential equation is

2 2 2 2

2 2 2 2

8 m(E V) 0

x y z h

where x, y, z are certain coordinates of the electron, m = mass of the

electron E = total energy of the electron. V = potential energy of the

electron; h = Planck‟s constant and (psi) = wave function of the electron.

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Significance of : The wave function may be regarded as the amplitude

function expressed in terms of coordinates x, y and z. The wave function

may have positive or negative values depending upon the value of

coordinates. The main aim of Schrodinger equation is to give solution for

probability approach. When the equation is solved, it is observed that for

some regions of space the value of is negative. But the probability must

be always positive and cannot be negative, it is thus, proper to use 2 in

favour of .

Significance of 2: 2 is a probability factor. It describes the probability of

finding an electron within a small space. The space in which there is

maximum probability of finding an electron is termed as orbital. The

important point of the solution of the wave equation is that it provides a set

of numbers called quantum numbers which describe energies of the electron

in atoms, information about the shapes and orientations of the most

probable distribution of electrons around nucleus.

Nodal Points and Planes:

The point where there is zero probability of finding the electron is called

nodal point. There are two types of nodes: Radial nodes and angular nodes.

The former is concerned with distance from the nucleus while latter is

concerned with direction.

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No. of radial nodes = n – – 1

No. of angular nodes =

Total number of nodes = n – 1

Nodal planes are the planes of zero probability of finding the electron. The

number of such planes is also equal to .

Illustration 46:

Calculate radial nodes and angular nodes for the following type of

orbitals.

(a) 1s (b) 2p

(c) 3p (d) 3d

(e) 4s (f) 4d

Solution:

(a) 0, 0 (b) 0, 1

(c) 1, 1 (d) 0, 2

(e) 3, 0 (f) 1, 2

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PHOTOELECTRIC EFFECT

Sir J.J. Thomson, observed that when a light of certain frequency strikes the

surface of a metal, electrons are ejected from the metal. This phenomenon

is known as photoelectric effect and the ejected electrons are called

photoelectrons.

A few metals, which are having low ionization energy like Cesium, show this

effect under the action of visible light but many more show it under the

action of more energetic ultraviolet light.

V

A

–+electrons

Evacuated quartz tube

Light

An evacuated tube contains two electrodes connected to a source of variable

voltage, with the metal plate whose surface is irradiated as the anode. Some

of the photoelectrons that emerge from this surface have enough energy to

reach the cathode despite its negative polarity, and they constitute the

measured current. The slower photoelectrons are repelled before they get to

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the cathode. When the voltage is increased to a certain value V0, of the

order of several volts, no more photoelectrons arrive, as indicated by the

current dropping to zero. This extinction voltage (or also referred as

stopping potential) corresponds to the maximum photoelectron kinetic

energy i.e., eVo = ½ mv2

The experimental findings are summarised as below:

Electrons come out as soon as the light (of sufficient energy) strikes

the metal surface.

The light of any frequency will not be able to cause ejection of

electrons from a metal surface. There is a minimum frequency, called

the threshold (or critical) frequency, which can just cause the ejection.

This frequency varies with the nature of the metal. The higher the

frequency of the light, the more energy the photoelectrons have. Blue

light results in faster electrons than red light.

Photoelectric current is increased with increase in intensity of light of

same frequency, if emission is permitted i.e., a bright light yields more

photoelectrons than a dim one of the same frequency, but the electron

energies remain the same.

Light must have stream of energy particles or quanta of energy (h ).

Suppose, the threshold frequency of light required to eject electrons from a

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metal is 0, when a photon of light of this frequency strikes a metal it

imparts its entire energy (h 0) to the electron.

E = h 0 E > h 0

K.Emax = h – h 0 K. E = 0

Metal

“This energy enables the electron to break away from the atom by

overcoming the attractive influence of the nucleus”. Thus each photon can

eject one electron. If the frequency of light is less than 0 there is no

ejection of electron. If the frequency of light is higher than 0 (let it be ),

the photon of this light having higher energy (h ), will impart some energy

to the electron that is needed to remove it from the atom. The excess

energy would give a certain velocity (i.e, kinetic energy) to the electron.

h = h 0 + K.E

h = h 0 + ½ mv2

½ mv2 = h –h o

where, = frequency of the incident light

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0 = threshold frequency

h 0 is the threshold energy (or) the work function denoted by = h 0

(minimum energy of the photon to liberate electron). It is constant for

particular metal and is also equal to the ionization potential of gaseous

atoms.

The kinetic energy of the photoelectrons increases linearly with the

frequency of incident light. Thus, if the energy of the ejected electrons is

plotted as a function of frequency, it result in a straight line whose slope is

equal to Planck‟s constant „h‟ and whose intercept is h 0.

K.Eof

Photoelectrons

Illustration 47:

Work function of sodium is 2.5 eV. Predict whether the wavelength

6500o

A is suitable for a photoelectron ejection or not.

Solution:

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Energy of incident light 34 8

10

hc 6.62 10 5 10

6500 10

= 3.055 10 19 J

= 1.9 eV

Which is lower than work function. Hence no ejection will take place.

Exercise 6:

(i) Find the threshold wavelengths for photoelectric effect from a copper

surface, a sodium surface and a caesium surface. The work function

of these metal are 4.5 eV, 2.3 eV and 1.9 eV respectively.

(ii) Energy required to stop the ejection of electrons from Cu plate is 0.24

eV. Calculate the work function when radiation of = 253.7 nm

strikes the plate?

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ANSWER TO EXERCISES

Exercise 1:

(i) 19

(ii) (a) 1.2044 1022

(b) 2.01 10–5 kg

Exercise 2:

(i) 1 2 2

2 1 1

E hc 2000 1

E hc 6000 3

(ii) 15 7 1v 3 10 Hz, 10 m

Exercise 3:

(i) 6603 Å

(ii) 9.84 105 J mol–1

(iii) 486 nm, Blue

Exercise 4:

(i) (a) 2 2 6 2 6 31s 2s 2p 3s 3p 3d

(b) 2 2 6 2 6 51s 2s 2p 3s 3p 3d

(c) 2 2 6 2 6 10 11s 2s 2p 3s 3p 3d 4s

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(ii) (a) 4d

(b) 2p

(c) 3s

Exercise 5:

(i) 10–7 m

(ii) 3.87 10–11 m

(iii) (a) 5.27 10–28m

(b)

Exercise 6:

(i) 276 nm, 540 nm, 654 nm

(ii) 4.65 eV

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MISCELLANEOUS EXERCISES

Exercise 1:

Calculate the ratio of specific charge (e/m) of a proton and that of an

-particle.

Exercise 2:

What is the fraction of volume occupied by the nucleus with respect to

the total volume of an atom?

Exercise 3:

Calculate the minimum and maximum values of wavelength in Balmer

series of a H atom.

Exercise 4:

The wave number of the first line of Balmer series of hydrogen is

15,200 cm 1. What is the wave number of the first line of Balmer

series for the Li2+ ion?

Exercise 5:

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What possible can be the ratio of the de Broglie wavelengths for two

electrons having the same initial energy and accelerated through 50

volts and 200 volts?

Exercise 6:

What is likely to be the principle quantum number for a circular orbit of

diameter 20 nm of the hydrogen atom if we assume Bohr orbit to be

the same as that represented by the principle quantum number?

Exercise 7:

Calculate the uncertainty in the velocity of a particle weighing 25.0 g if

the uncertainty in position is 10 5 m. (given Planck constant h = 6.6

10 34 Js).

Exercise 8:

Calculate the de Broglie wavelength of a tennis ball of mass 60.0 g

moving with a velocity of 10 metres per second (Planck constant, h =

6.63 10 34 Js).

Exercise 9:

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Suppose 10–17 J of energy is needed by the interior of human eye to

see an object. How many photons of green light ( = 550 nm) are

needed to generate this minimum amount of energy?

Exercise 10:

Ionization potential of hydrogen atom 13.6 eV. If hydrogen atom in

ground state excited by monochromatic light of energy 12.1 eV, then

what will be the total spectral lines emitted according to Bohr‟s theory?

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ANSWERS TO MISCELLANEOUS EXERCISES

Exercise 1:

1 : 2

Exercise 2:

10 15

Exercise 3:

3647 nm and 6564 nm

Exercise 4:

1,36,800 cm 1

Exercise 5:

2 : 1

Exercise 6:

14

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Exercise 7:

2.1 10 28

Exercise 8:

10 33 m

Exercise 9:

28

Exercise 10:

3

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SOLVED PROBLEMS

Subjective:

Board Type Questions

Problem 1:

Why are Bohr‟s orbits are called stationary states?

Solution:

This is because the energies of orbits in which the electrons revolve

are fixed.

Problem 2:

Explain why the electronic configuration of Cu is 3d104s1 and not

3d94s2.

Solution:

In the 3d104s1 the d-subshell is completely filled which is more stable.

Problem 3:

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Fe3+ ion is more stable than Fe2+ ion. Why?

Solution:

In Fe3+ ion 3d subshell is half filled hence more stable configuration.

Problem 4:

Calculate the accelerating potential that must be applied to a proton

beam to give it an effective wavelength of 0.005 nm.

Solution:

hv

m

21eV mv

2

Putting the values we get

V = 32.85 volt

Problem 5:

Give one example of isodiapheres.

Solution:

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Isodiapheres have same difference between the number of neutrons

and protons. For example

39 31

19 15n p 1 n p 1

K & P

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IIT Level Questions

Problem 6:

Which electronic transition in Balmer series of hydrogen atom has

same frequency as that of n = 6 to n = 4 transition in He+. [Neglect

reduced mass effect].

Solution:

2He

2 2

1 1RZ

4 6

= 36 16

4R36 16

= 5R

36

H = R 122 2

1 1

2 n

HHev v

5R

36=

2

R R

4 n

On solving above equation

n2 = 9

n = 3

Or corresponding transition from 3 2 in Balmer series of hydrogen

atom has same frequency as that of 6 4 transition in He+.

Problem 7:

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Calculate ionization potential in volts of (a) He+ and (b) Li2+

Solution:

I.E. = 2

2

13.6Z

n

= 2

2

13.6 Z

1 [Z =2 for He+]

= 13.6 4 = 54.4 eV

Similarly for Li2+ = 2

2

13.6 3

1

= 13.6 9 = 122.4 eV

Problem 8:

Calculate the ratio of K.E and P.E of an electron in an orbit?

Solution:

K.E. = 2Ze

2r

P.E. = 2Ze

r

P.E. = –2K.E

K.E 1

=-P.E 2

Problem 9:

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How many spectral lines are emitted by atomic hydrogen excited to

nth energy level?

Solution:

1 1+2=3n = 1

n = 2

n = 3

n = 4

n = 5

n = 6

1+2+3= 6

1

2

3

1

2

3

4

3

1

5 6

Thus the number of lines emitted from nth energy level

= 1 + 2 + 3 +………… n – 1 = (n – 1)

n =n n 1

2

(n – 1) = n 1 n 1 1

2=

n 1 n

2

Number of spectral lines that appear in hydrogen spectrum when an

electron jumps from nth energy level = n n-1

2

Problem 10:

Calculate (a) the de Broglie wavelength of an electron moving with a

velocity of 5.0 105 ms–1 and (b) relative de Broglie wavelength of an

atom of hydrogen and atom of oxygen moving with the same velocity

(h = 6.63 10–34 kg m2 s–1)

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Solution:

(a) = h

mv =

34 2 2

31 5 1

6.63 10 kgm s

9.11 10 kg 5.0 10 ms

Wavelength = 1.46 10–9m

(b) An atom of oxygen has approximately 16 times the mass of an

atom of hydrogen. In the formulah

mv, h is constant while the

conditions of problem make v, also constant. This means that

and m are variables and varies inversely with m. Therefore,

for the hydrogen atom would be 16 times greater than for

oxygen atom.

Problem 11:

A 1 MeV proton is sent against a gold leaf (Z = 79). Calculate the

distance of closest approach for head on collision.

Solution:

K.E. = P.E.

2

2

0

1 1 Zemv

2 4 d

Hence d = 2

2

0

Ze

14 mv

2

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29 19

13

13

9 10 79 1.6 10as 1MeV 1.6 10 J

1.6 10

131.137 10 m

Problem 12:

What is the wavelength associated with 150 eV electron

Solution:

= h

2 m K.E.

= 34

31 19

6.626 10 Js

2 9.1 10 kg 150 1.6 10 J =

34

50

6.626 10

4368 10 = 10–10 m = 1 Å

Problem 13:

The energy of electron in the second and third Bohr orbit of the

hydrogen atom is –5.42 10–12 erg and –2.41 10–12erg, respectively.

Calculate the wavelengths of emitted radiation when the electron

drops from third to second orbit.

Solution:

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E3 – E2 = h = hc

– 2.41 10–12 – (– 5.42 10–12) =

27 106.626 10 3 10

= 27 10

12

6.626 10 3 10

3.01 10

= 6.604 10–5 cm = 6.604 10–5 108 = 6604 Å

Problem 14:

O2 undergoes photochemical dissociation into one normal oxygen and

one excited oxygen atom, 1.967 eV more energetic than normal. The

dissociation of O2 into two normal atoms of oxygen atoms requires

498kJ mole–1. What is the maximum wavelength effective for

photochemical dissociation of O2?

Solution:

O2 ON + excitedO

O2 ON + ON

E = 498 103 J / mole

=3

23

498 10J

6.023 10 per molecule = 8.268 10–19 J

Energy required for excitation = 1.967 eV = 3.146 10–19J

Total energy required for photochemical dissociation of O2

= 8.268 10–19 + 3.146 10–19 = 11.414 10–19 J hc

= 11.414 10–19J

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= 34 8

19

6.626 10 3 10

11.414 10= 1.7415 10–7 m = 1741.5 Å

Problem 15:

Compare the wavelengths for the first three lines in the Balmer series

with those which arise from similar transition in Be3+ ion. (Neglect

reduced mass effect).

Solution:

2

H 2 2

1 1v R 1

2 n

2

Be 2 2

1 1v R 4

2 n

Be H

H Be

v

v = 16

So we can conclude that all transitions in Be3+ will occur at

wavelengths 1

16times the hydrogen wavelengths.

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Objective:

Problem 1:

For a p-electron, orbital angular moment is

(A) 2 (B)

(C) 6 (D) 2

Solution:

Orbital angular momentum L = ( 1) where h

2

L for p electron = 1(1 1) 2

(A)

Problem 2:

For which of the following species, Bohr theory doesn‟t apply

(A) H (B) He+

(C) Li2+ (D) Na+

Solution:

Bohr theory is not applicable to multi electron species

(D)

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Problem 3:

If the radius of 2nd Bohr orbit of hydrogen atom is r2. The radius of

third Bohr orbit will be

(A) 2

4r

9 (B) 4r2

(C) 2

9r

4 (D) 9r2

Solution:

2 2

2 2

n hr

4 mZe

2

2

2

3

r 2

r 3

r3 = 2

9r

4

(C)

Problem 4:

Number of waves made by an electron in one complete revolution in

3rd Bohr orbit is

(A) 2 (B) 3

(C) 4 (D) 1

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Solution:

Circumference of 3rd orbit = 2 r3

According to Bohr‟s angular momentum of electron in 3rd orbit is

mvr3 = h

32

or 32 rh

mv 3

By de-Broglie equation,

= h

mv

= 32 r

3

2 r3 = 3

i.e. circumference of 3rd orbit is three times the wavelength of electron

or number of waves made by Bohr electron in one complete

revolution in 3rd orbit is three.

(B)

Problem 5:

The degeneracy of the level of hydrogen atom that has energy HR-16

is

(A) 16 (B) 4

(C) 2 (D) 1

Solution:

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En = H

2

R

n

H H

2

R R

16n

i.e. for 4th sub-shell

- 1 0 +1 - 2 –1 +1 0 +2 – 3 – 2 – 1 0 +1 +2 +3

3 2 1

three p five d Seven f

n = 4 l = 0

m = 0 one s

i.e. 1 + 3 + 5 + 7 = 16

Degeneracy is 16

Problem 6:

An electron is moving with a kinetic energy of 4.55 10–25 J. What will

be de Broglie wavelength for this electron?

(A) 5.28 10–7 m (B) 7.28 10–7 m

(C) 2 10–10 m (D) 3 10–5 m

Solution:

KE = 1

2 mv2 = 4.55 10–25

v2 = 25

6

31

2 4.55 101 10

9.1 10

v = 103 m/s

de Broglie wavelength = h

mv =

34

31 3

6.626 10

9.1 10 10 = 7.28 10–7 m

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(B)

Problem 7:

Suppose 10–17J of energy is needed by the interior of human eye to

see an object. How many photons of green light ( = 550 nm) are

needed to generate this minimum amount of energy?

(A) 14 (B) 28

(C) 39 (D) 42

Solution:

Let the number of photons required = n

n 17hc10

n = 1710

hc =

17 9

34 8

10 550 10

6.626 10 3 10 = 27.6 = 28 photons

(B)

Problem 8:

The two electrons present in an orbital are distinguished by

(A) principal quantum number (B) azimuthal quantum number

(C) magnetic quantum number (D) spin quantum number

Solution:

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They are distinguished by their spin.

(D)

Problem 9:

The velocity of electron in the ground state hydrogen atom is

2.18 106 ms–1. Its velocity in the second orbit would be

(A) 1.09 106 ms–1 (B) 4.38 106 ms–1

(C) 5.5 105 ms–1 (D) 8.76 106 ms–1-

Solution:

We know that velocity of electron in nth Bohr‟s orbit is given by

= 2.18 106 Z

n m/s

For H, Z = 1

v1 = 62.18 10

1 m/s

v2 = 62.18 10

2 m/s = 1.09 106 m/s

(A)

Problem 10:

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The ionization energy of the ground state hydrogen atom is

2.18 10–18J. The energy of an electron in its second orbit would be

(A) –1.09 10–18 J (B) –2.18 10–18J

(C) –4.36 10–18J (D) –5.45 10–19J

Solution:

Energy of electron in first Bohr‟s orbit of H–atom

E = 18

2

2.18 10J

n ( ionization energy of H = 2.18 10–18J)

E2 = 18

2

2.18 10

2 J = –5.45 10–19J

(D)

True and False

Problem 11:

All electromagnetic radiations have same energy.

Solution:

False.

Different electromagnetic radiation have different wavelengths, hence

different energy

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Problem 12:

Bohr‟s model is applicable for H-atom only.

Solution:

False.

Bohr‟s model is applicable for species having one electron like H, He+,

Li2+.

Problem 13:

Specific charge of cathode rays remains same irrespective of the gas

used in the discharge tube.

Solution:

True.

Cathode rays contain electrons which remains same with every gas.

Problem 14:

Isoelectronic species have same electronic configuration.

Solution:

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True.

Isoelectronic species have same number of electrons.

Problem 15:

Atoms with same atomic number but different number of neutrons are

called isotopes.

Solution:

True.

Isotopes have same atomic number but different atomic mass.

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Fill in the Blanks

Problem 16:

The mass of positron is ___________ electron.

Solution:

Equal to

Problem 17:

With increasing principle quantum number, the energy difference

between adjacent energy levels in H-atom _________.

Solution:

Decreases

Problem 18:

If the magnetic moment of an ion An+ is 5.9, then the number of

unpaired electrons present in An+ are ____________.

Solution:

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Five. Applying = n n 2 , where n 5

Problem 19:

s-electrons are more penetrating than p-electrons. The energy

required to abstract s-electrons is ____________ than for p electrons.

Solution:

Greater

Problem 20:

The wavelength of the first spectral line in Paschen series for H-atom

is____________.

Solution:

16410 Å

For Paschen series 2 1n 4, n 3

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ASSIGNMENT PROBLEMS

Subjective:

Level – O

1. How many electrons in an atom have the following quantum numbers?

(i) n = 4, ms = 1/2 (ii) n = 3, 0

2. Write the number of protons, electrons and neutrons in nitride ion.

3. Distinguish between absorption and emission spectra.

4. Write down the electronic configuration of Mn4+ and Cr3+. How many

unpaired electrons are present in them?

5. Write correct orbital notation for each of the following sets of quantum

numbers.

(i) n = 1, 0 , m = 0 (ii) n = 2, 1 , m = 1

6. What are drawbacks of Rutherford‟s model atom?

7. Explain why uncertainty principle is significant only for the motion of

subatomic particles but not for macroscopic objects.

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8. Why we do not see a car moving in the form of a wave on a road?

9. Why e/m ratio in case of anode rays depends upon the type of gas?

10. Why splitting of spectral lines take place when the source giving the

spectrum is placed in a magnetic field?

11. What is the maximum number of lines obtained when the excited

electrons of a hydrogen atom in n = 6 drops to the ground state.

12. Calculate the radius of the third orbit of a hydrogen atom.

13. The wavelength of a moving body of mass 0.1 mg is 293.31 10 m.

Calculate its kinetic energy 34h 6.625 10 J sec

14. Calculate the mass and charge of one mole of electrons.

15. Calculate (a) wave number and (b) frequency of yellow radiations

having wavelength of 5800o

A .

16. In one discharge tube H2 gas is taken and in other O2 gas is taken. Will

the negative and positive charged particles in cathode rays and anode

rays be same?

17. Explain why half filled and full filled configurations are more stable.

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18. Some energy is absorbed by hydrogen atom due to which an electron

in it jumped from ground state to the state having principle quantum

number 5 and again jumped back to the original level. What type of

spectrum is observed and in which region. What is the name of series?

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Level – I

1. What is the mass of a photon of sodium light with a wavelength of

5890o

A ? 27 10 1h 6.63 10 ergsec,c 3 10 cmsec

2. Calculate the kinetic energy of a moving electron which has a

wavelength of 4.8 pm (Mass of electron 31 349.11 10 kg,h 6.63 10 Js)

3. Calculate de – Broglie wavelength of an electron moving with 1% of

the speed of light.

4. Calculate the energy of an electron in the second Bohr orbit of an

excited hydrogen atom

5. Calculate the velocity of an electron revolving in the second orbit of a

hydrogen atom

6. How many photons of light having a wavelength of 4000 o

A are

necessary to provide 1 J of energy? (h = 6.63 10-34Js, c = 3 108 m/s)

7. How the electron in the ground state of hydrogen atom is excited by

means of monochromatic radiation of wavelength 970.6o

A . How many

different lines are possible in the resulting emission spectrum? Find the

longest wavelength among these?

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8. The wavelength of the first line in the Balmer series is 656 nm.

Calculate the wavelength of the second line and the limiting line in

Balmer series.

9. A single electron system has ionization energy 11180 kJ / mol. Find

the number of protons in the nucleus of the system.

10. Calculate the velocity of an electron placed in the third orbit of

hydrogen atom. Also calculate the number of revolutions per second

that this electron makes around the nucleus.

11. (a) What is wavelength of a particle of mass 1 g moving with a

velocity of 200 m/s?

(b) A moving electron has 4.55 10-25 joules of kinetic energy.

Calculate its wavelength (mass = 9.1 10–31 kg and h = 6.6 10–

34 kg m2sec–1)

(c) Calculate the de-Broglie wavelength of electron accelerated

through 100 volt.

12. Calculate the radii of the I, II and III permitted electron Bohr orbits in

a hydrogen atom. What are the corresponding values in the case of a

singly ionised helium atom?

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13. Calculate energy in kcal/mole necessary to remove an electron in a

hydrogen atom in fourth principal quantum number to infinity.

14. A monochromatic source of light operating at 600 watt emits 2 1022

photons per second. Find the wavelength of the light.

15. Calculate the uncertainty in the position of an electron if it has a speed

of 500 m/s with an uncertainity of 0.02%.

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Level – II

1. (a) The electron energy in hydrogen atom is given by E = – 21.7

10–12/n2 erg. Calculate the energy required to remove an

electron completely from n = 2 orbit. What is the longest

wavelength (in cm) of light that can be used to cause this

transition?

(b) Ionization energy of hydrogen atom is 13.6 eV. Calculate the

ionization energy for Li2+ and Be3+ in the first excited state.

2. A bulb emits light of wave length 4500 Å. The bulb is rated as 150

watt and 8% of the energy is emitted as light. How many photons are

emitted by bulb per second?

3. (a) What Amount of accelerating potential is needed to produce an

electron beam with an effective wavelength of 0.09 Å.

(b) Calculate de-Broglie wavelength of an electron moving with a

speed of nearly 1

20th that of light (3 108 ms–1)

4. A doubly ionised lithium atom is hydrogen like with an atomic

no. 3.

(a) Find the wavelength of the radiation required to excite the

electron in Li from the first to the third Bohr orbit (ionisation

energy of the hydrogen atom is equal to 13.6 eV)

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(b) How many spectral lines are observed in the emission spectrum

of the above excited system?

5. Calculate of the radiations when the electron jumps from III to II

orbit of hydrogen atom. The electronic energy in II and III Bohr orbit

of hydrogen atoms are – 5.42 10–12 and –2.41 10–12erg respectively.

6. 1.8 g hydrogen atoms are excited to radiations. The study of spectra

indicates that 27% of the atoms are in 3rd energy level and 15% of

atoms in 2nd energy level and the rest in ground state. IP of H is 13.6

eV. Calculate (i) No. of atoms present in III and II energy level (ii)

Total energy evolved when all the atoms return to ground state.

7. The IP of H is 13.6 eV. It is exposed to electromagnetic waves of 1028

Å and given out induced radiations. Find the wavelength of these

induced radiations.

8. What is uncertainty in velocity of an electron if uncertainty in its

position is 1Å?

9. A single electron beam, atom has nuclear charge +Ze where Z is

atomic number and e is electronic charge. It requires 16.52 eV to

excite the electron from the second Bohr orbit to third Bohr orbit. Find

(a) The atomic no. of element

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(b) The energy required for transition of electron from first to third

orbit.

(c) Wavelength required to remove electron from first Bohr orbit to

infinity.

(d) The kinetic energy of electron in first Bohr orbit.

10. A sample of hydrogen gas containing some atoms in one excited state

emitted three different types of photons. When the sample was

exposed with radiation of energy 2.88 eV it emitted 10 different types

of photons, all having energy equal or less than 13.05 eV. Find out

(a) The principal quantum numbers of initially excited electrons.

(b) The principal quantum numbers of electrons in final excited

state.

(c) The maximum and minimum energies of initially emitted

photons.

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Objective:

Level – I

1. The ratio of energy of the electron in ground state of the hydrogen to

electron in first excited state of He+ is

(A) 1 : 4 (B) 1 : 1

(C) 1 : 8 (D) 1 : 16

2. Bohr model can explain spectrum of

(A) the hydrogen atom only

(B) all elements

(C) any atomic or ionic species having one electron only

(D) the hydrogen molecule

3. Which is the correct order of probability of being found close to the

nucleus is

(A) s p d f (B) f d p s

(C) p d f s (D) d f p s

4. Ratio between longest wavelengths of H atom in Lyman series to the

shortest wavelength in Balmer series of He+ is

(A) 4

3 (B)

36

5

(C) 1

4 (D)

5

9

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5. If the radius of first Bohr orbit is a, then de-Broglie wavelength of

electron in 3rd orbit is nearly.

(A) 2 a1 (B) 6 a1

(C) 9 a1 (D) 16 a1

6. If uncertainty in position and momentum are equal, the uncertainty in

velocity would be

(A) 1 h

2m (B)

h

2

(C) h

(D) h

m

7. Which of the following is a coloured ion?

(A) Cu+(aq) (B) Na+(aq)

(C) Cu2+ (aq) (D) K+ (aq)

8. The number of orbitals in a sub-shell are given by

(A) 2 (B) n2

(C) 2 + 1 (D) 2n2

9. For a „d‟ electron, the orbital angular momentum is

(A) 6h/2 (B) 2h/2

(C) h/2 (D) 2h/2

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10. Hydrogen atom consists of a single electron but so many lines appear

in the spectrum of atomic hydrogen because

(A) Sample contains some impurity

(B) Experiment is done on collection of atoms.

(C) Hydrogen atom splits to form more than one different species.

(D) Some different isotope of hydrogen atom may be present.

11. A line with wave number 1.028 10-2 nm-1 is emitted in the spectrum of

atomic hydrogen. In what region of the electromagnetic spectrum does

this line occur?

(A) far UV (B) near UV

(C) near IR (D) far IR

12. One molecule of a substance absorbs one quantum of energy. The

energy involved when 1.5 mole of the substance absorbs red light of

frequency 7.5 1014 sec–1 will be

(A) 2.99 105 J (B) 3.23 105J

(C) 4.48 105J (D) 2.99 106 J

13. If a shell is having g sub-shell, which is correct statement about

principal quantum number n of this shell.

(A) n 5 (B) n 5

(C) n = 5 (D) Cannot be determined

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14. The wave number of a spectral line is 5 105 m–1. The energy

corresponding to this line will be

(A) 3.39 10–23 kJ (B) 9.93 10–23 kJ

(C) 3.45 10–24 J (D) none of these

15. Wave length of the radiation when electron jumps from second shell to

1st shell of H atom (RH = 109679 cm–1)

(A) 1215.6 Å (B) 1397.5 Å

(C) 2395.87 Å (D) none of these

16. The orbital which have lobe along the axis is/are

(A) 2zd (B) dxz

(C) 2 2x yd (D) dxy

17. Anode rays

(A) contain positively charged particles

(B) have constant e/m ratio irrespective of the type of gas used

(C) may contain He+ ions

(D) are always beams of protons

18. The spectrum of He is expected to be similar to that of

(A) H (B) Li+

(C) Be2+ (D) H+

19. The total energy of electron in an atom

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(A) is less than zero

(B) is the sum of kinetic energy and potential energy

(C) is equal to kinetic energy in magnitude

(D) is equal to potential energy in magnitude

20. Which of the following orbitals have no nodal plane?

(A) px (B) dxy

(C) 2zd (D) 4s

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Assertion–Reason type questions

The following questions consist of two statements each printed as Assertion

and Reason. While answering these questions, you are required to choose

any one of the following five responses.

(A) If both Assertion and Reason are true and the Reason is a correct

explanation of the Assertion.

(B) If both Assertion and Reason are true but Reason is not a correct

explanation of the Assertion.

(C) If Assertion is true but the Reason is false.

(D) If the Assertion is false but the Reason is true.

(E) If both Assertion and Reason are false.

21. A: A spectral line will be observed for 2px-2py transition.

R: The energy will be released in the form of electromagnetic

radiations.

22. A: Cs is used in photoelectric cells.

R: Cs is an alkali metal.

23. A: The energy of an electron is largely determined by its principle

quantum number.

R: 2s subshell has higher energy than 1s.

24. A: Fe3+(g) ion is more stable than Fe2+(g).

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R: Fe3+(g) ion has higher magnetic moment than Fe2+(g) ion.

25. A: de-Broglie‟s equation and uncertainity principle are applicable to

moving bodies.

R: These have significance to microscopic as well as macroscopic

particles.

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Level - II

1. Which statement is wrong about Bohr‟s theory?

(A) Orbit is a three dimensional area where probability of finding

electron is maximum.

(B) Orbit is a two dimensional track on which electron moves

(C) Atom has definite boundary

(D) Energies and angular momentum of orbits are quantized.

2. The introduction of a neutron into the nuclear composition of an atom

would lead to a change in

(A) The number of electrons also

(B) The chemical nature of the atom

(C) Its atomic number

(D) Its atomic weight

3. Which statement is true?

(A) Spacing between energy levels n = 1 and n = 2 in hydrogen

atom is greater than that of n = 2 and n = 3

(B) Spacing between energy levels n = 1 and n = 2 in hydrogen

atom is equal to that n = 2 and n = 3

(C) Spacing between energy levels n = 1 and n = 3 in hydrogen

atom is less than that of n = 2 and n = 3

(D) None of these

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4. If the wavelength of first line of the Balmer series of hydrogen atom is

656.1 nm, the wavelength of second line of this series would be

(A) 218.7 nm (B) 328.0 nm

(C) 86.0 nm (D) 640.0 nm

5. In absence of external magnetic field f sub-shell is

(A) 5 fold degenerate (B) 3 fold degenerate

(C) 7 fold degenerate (D) non-degenerate

6. Which set of quantum numbers is not possible for electron in 3rd shell?

(A) n = 3, = 2, m = – 1, s = +1/2

(B) n = 3, = 2, m = – 1, s = –1/2

(C) n = 3, = 2, m = 0, s = +1/2

(D) n = 3, = 3, m = 0, s = –1/2

7. The first four ionization energies of an element are 191, 578,872 and

5962 kcal. The number of valence electrons in the element is

(A) 1 (B) 2

(C) 3 (D) 4

8. The radiation is emitted when a hydrogen atom goes from a high

energy state to a lower energy state. The wavelength of one line in

visible region of atomic spectrum of hydrogen is 6.5 10–9 m. Energy

difference between the two states is

(A) 3.0 10–17 J (B) 1.0 10–18 J

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(C) 5.0 10–10 J (D) 6.5 10–7 J

9. The ratio of the energy of the electron in ground state of hydrogen to

the electron in first excited state of Be3+ is

(A) 1 : 4 (B) 1 : 8

(C) 1 : 16 (D) 16 : 1

10. Which one of the following species is isoelectronic with P3–?

(A) Kr (B) Ca2+

(C) Na+ (D) F–

11. Which of the following sets of quantum numbers is/are not allowable?

(A) n = 3, = 2, m = 0

(B) n = 2, = 0, m = – 1

(C) n = 4, = 3, m = + 1

(D) n = 1, = 0, m = 0

12. Among V (Z = 23), Cr (Z = 24), Mn (Z = 25) which will have highest

magnetic moment.

(A) V

(B) Cr

(C) Mn

(D) all of them will have equal magnetic moment

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13. If velocity of an electron in 1st Bohr orbit of hydrogen atom is x, its

velocity in 3rd orbit will be

(A) x

3 (B) 3x

(C) 9x (D) x

9

14. Which element has a hydrogen like spectrum whose lines have

wavelengths one fourth of atomic hydrogen?

(A) He+ (B) Li 2+

(C) Be3+ (D) B4+

15. Calculate the wavelength of a track star running 150 metre dash in

12.1 sec if its weight is 50 kg.

(A) 9.11 10–34 m (B) 8.92 10–37 m

(C) 1.12 10–45 metre (D) none of these

16. Which of the following sets of quantum numbers is not allowed

(A) n = 3, = 1, m = +2 (B) n = 3, = 1, m = +1

(C) n = 3, = 0, m = 0 (D) n = 3, = 2, m = 2

17. Assuming that a 25 watt bulb emits monochromatic yellow light of

wavelength 0.57 . The rate of emission of quanta per sec. will be

(A) 5.89 1015 sec–1 (B) 7.28 1017 sec–1

(C) 5 1010 sec–1 (D) 7.18 1019 sec–1

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18. How many chlorine atoms can you ionize in the process Cl Cl + e,

by the energy liberated from the following process?

Cl + e– Cl for 6 1023 atoms

Given electron affinity of Cl = 3.61 eV, and I P of Cl = 17.422 eV

(A) 1.24 1023 atoms (B) 9.82 1020 atoms

(C) 2.02 1015 atoms (D) none of these

19. The binding energy of an electron in the ground state of the He atom

is equal to 24 eV. The energy required to remove both the electrons

from the atom will be

(A) 59 eV (B) 81 eV

(C) 79 eV (D) None of these

20. The wave number of the shortest wave length transition in Balmer

series of atomic hydrogen will be

(A) 4215 Å (B) 1437 Å

(C) 3942 Å (D) 3647 Å

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ANSWERS TO ASSIGNMENT PROBLEMS

Subjective:

Level – O

1. (i) Total electrons in n = 4 are 2n2 i.e. 2 42 = 32. Half of them

have ms = 1/2.

(ii) n = 3, 0 means 3s orbitals which can have 2 electrons.

2. Nitride ion is N3 .

P = 7, e = 10, n = 7

3. Refer to theory.

4. Mn4+ (Z = 25) = 1s22s2sp63s23p63d3. It has three unpaired electrons.

Cr3+ (Z = 24) = 2s22s22p63s23p63d3. It also has three electrons.

5. (i) 1s (ii) 2px or 2py

6. Refer to theory.

8. We know that = h/mv, since mass of the car is large hence

wavelength is very small.

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9. Anode rays are beams of positive ions which are different for different

gases.

10. In the presence of magnetic field the orbitals present in a subshell take

up different orientations.

11. Number of lines n n 1 6 6 1

152 2

12. 4.77 10 8

13. 2 10 3

14. Mass = 5.486 107 kg and charge = 9.65 104 C

15. (a) 6 11.72 10 m

(b) 145.172 10 Hz

16. Cathode rays are same in both cases, but anode rays will be different.

17. Refer to theory.

18. Emission spectrum, Lyman series.

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Level – I

1. 333.75 10 g

2. 141.047 10 J

3. 102.4 10 mk 2.4A

4. 0.545 10-11 erg

5. 1.09 108 cm/sec

6. 2.01 1018

7. = 1215.68 A

8. 364.4 nm

9. Z = 3

10. V3 = 7.275 105 m/sec

rps = 2.432 1014

11. (a) 3.313 10–33m

(b) 7.25 10–7m

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(c) 1.227Å

12. r1= 0.53 Å : r1 = 0.265 Å

r2 = 2.12 Å for hydrogen : r2 = 1.06 Å for He+

r3 = 4.77 Å : r3 = 2.385 Å

13. 19.5 kcal/mole

14. 6630 nm

15. 5.77 10 4

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Level – II

1. (a) 3.67 10–5 cm

(b) 30.6 eV for Li2+, 54.4 eV for Be3+

2. 27.2 1018

3. (a) 1.86 104 eV

(b) 4.85 10–11m

4. (i) 113.7 Å

(ii) 3

5. 6603 Å

6. (i) 628.72 1021 atoms

(ii) 832.50 kJ.

7. 1 = 1028 Å, 2= 1216 Å, 3 = 6558 Å

8. 5.79 105

9. (i) 3, (ii) 108.87 eV, (iii) 1.01 10–6cm (iv) 122.4 eV

10. (a) 2 and 3

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(b) 5

(c) 12.09 eV and 1.9 eV

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Objective:

Level –I

1. A 2. C 3. A

4. A 5. B 6. A

7. C 8. C 9. A

10. B 11. A 12. C

13. B 14. B 15. A

16. A, C 17. A, C 18. B, C

19. A, B, C 20. C, D 21. D

22. B 23. B 24. B

25. C

Level – II

1. A 2. D 3. A

4. C 5. C 6. D

7. C 8. A 9. A

10. B 11. B 12. B

13. A 14. A 15. B

16. A 17. D 18. A

19. C 20. D

Atomic Structure Chemistry

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