Transcript
J 10 x 2.18R ),n
1(RE 18-
H2Hn
J 10 x 2.18R where),n
1
n
1(RΔE 18-
H2
f
2
i
H -
At the end of this lesson, students should be able to:
describe Bohr’s atomic model.
OBJECTIVES
explain the existence of energy levels in an atom.
calculate the energy of an electron using:
calculate the energy change of an electron during transition.
calculate the photon emitted by an electron that produces a
particular wavelength during transition.
DALTON ------ ATOMS
THOMSON ------ ELECTRONS
RUTHERFORD ------- POSITIVELY CHARGED NUCLEUS
Four Bohr’s atomic postulate :
electron moves in circular orbit about the nucleus.
Each orbit has a different energy level/energy state. 1
In the specific energy level, the energy of electron is
fixed in value or is quantised.
The electron does not release or absorb any energy
when it is in the ground state or at any particular
energy level/orbit.
This means that an electron moving in an orbit can
have a certain amount of energy.
(quantised = fixed/specific/definite)
2
2
1
nRE Hn
electron
nucleus
(proton)
n=1
n=2
energy of electron at n=1 ≠ energy of electron at n=2
The energy of an electron in its level is given by:
2
1
nRE Hn
where,
RH (Rydberg constant)= 2.18 10-18J. n (orbit/ energy level) = 1, 2, 3,…, (integer)
NOTES:
Energy is zero (E = 0) if electron is located infinitely
(n = )far from nucleus.
Energy associated with forces of attraction are taken to
be negative (thus, negative sign)
n=2 (higher energy level)
n=3 (higher energy level)
nucleus (proton)
n=1
3
energy absorbed
At ordinary condition, the electron is at the ground
state (lowest energy state).
If energy is supplied, electron absorbed the energy
and is promoted from lower energy level to a higher
energy level (electron is excited)
Energy at excited state is unstable.
It will fall back to lower energy level and released a
specific amount of energy in the form of light or
photon.
The light have specific wavelength. The line
spectrum is formed.
4
n=2 (higher energy level)
n=3 (higher energy level)
nucleus (proton)
n=1
energy released
The energy absorb / released during transition can be
calculated by:
22
11
fi
Hnn
RΔE
Where,
RH = Rydberg constant (2.18 x 10-18 J)
n= principal quantum number (1, 2, 3, …, )
ni = position(n) of electron initially
nf = position(n) of electron finally
The amount of energy released by the electron during
transition is called a photon of energy.
A photon of energy is released in the form of light with
appropriate frequency and wavelength.
Therefore, the energy change (released/absorbed) can
be related as:
Photon = a packet of light energy equals to h
hυΔE
where, h (Planck’s constant) =6.63 10-34 Js
= frequency; c (speed of light) = 3.00 108 ms-1
The energy of an electron in its level is given by:
2
1
nRE Hn
The energy absorb / released during transition can be
calculated by:
22
11
fi
Hnn
RΔE
hυΔE
A photon of energy is emitted in the form of radiation with
appropriate frequency and wavelength.
c
υWhere:
Thus:
hcΔE
Where:
v = frequency
λ = wavelength
c = speed of light (3.00x108 ms-1)
h = Plank’s constant (6.63 x 10-34 Js)
Energy of electron at particular orbit
2
1
nRE Hn
Calculation 1
where,
RH (Rydberg constant)= 2.18 10-18J.
n (orbit/ energy level) = 1, 2, 3,…, (integer)
Calculate the energy of electron that occupies n=2
Example 1
Answer:
The energy of an electron at its orbit (n) is -1.36x10-19J.
Determine the value of n.
Example 2
Answer:
1. Calculate the energy of electron that
occupies n=6
2. The energy of an electron at its orbit (n) is
-8.72x10-18J. Determine the value of n.
Energy released/absorbed by electron during transition
Calculation 2
Calculate the energy released when an electron falls from n=3
to n=2.
Example 1
Answer:
22
11
fi
Hnn
RΔE
RH = Rydberg constant (2.18 x 10-18 J)
n= principal quantum number (1, 2, 3, …, )
ni = position(n) of electron initially
nf = position(n) of electron finally
Calculate the energy required to promote an electron from the first
energy level to the third energy level of a hydrogen atom.
Example 2
J 10 x 1.94
3
1
1
11018.2
n
1
n
1RΔE
18
22
18
2
f
2
i
H
-
Jx
Answer:
1. Calculate the energy released when an electron excited from n=1 to n=4.
Energy released/absorbed by electron during transition
Calculation 3
Calculate the energy absorbed by an electron that produces a wavelength
of 9.38 x 10-8 m.
Example 1
Answer:
hυΔE
h (Planck’s constant) =6.63 10-34 Js
= frequency;
c (speed of light) = 3.00 108 ms-1
An electron absorbed 4.58x10-19 J of energy when falls from n=5 to n=2.
Calculate the frequency and wavelength produced.
Example 2
Answer:
1 14
34- 19-
10 x 6.91
Js)10 x (6.63 10 x 4.58
hυΔE
s
J
ms
smxc 7-
1 14
18
4.34x10 10 x 6.91
1000.3
cυ
released
1. Calculate the energy released by an electron that produces a wavelength of 9.49 x 10-8 m.
2. An electron released 2.09x10-18 J of energy when excited from n=1 to n=5. Calculate the frequency and wavelength produced.
<9.52 x 10-8m>
absorbed
Calculation 4
Calculate:
(a) The energy (in J) of an electron has when it occupies a
level equivalent to the quantum number of n = 3 and n = 4.
(b) The energy (in kJ/mol) of photon emitted when one mole of
electron drops from the 4th energy level to the 3rd energy
level.
(c) The frequency (in s-1) and wavelength (in nm) of this
photon.
Answer:
Calculate the energy of an electron in the hydrogen atom
when n = 2, and when n = 6.
Calculate the wavelength released when an electron
moves from n = 6 to n = 2.
If this line in the visible region of electromagnetic
spectrum?
(Note: visible light: 400 ~ 700 nm)
At the end of this lesson, students should be able to:
describe the formation of line spectrum of hydrogen atom.
OBJECTIVES
perform calculations involving the Rydberg equation for
Lyman, Balmer, Paschen, Brackett and Pfund series:
calculate the ionisation energy of hydrogen atom from
Lyman series.
n2 n1 and m 10 x 1.079R where
)n
1
n
1(R
1
1-7H
22
21
H -
State the weakness of Bohr’s atomic model.
State the dual nature of electron using de Broglie’s
postulate and Heisenberg’s uncertainty principle.
(calculation is not required)
Formation of line spectrum of hydrogen atom
Spectrum is a series of lines or a set of colours. 1
Line spectrum Continuous spectrum
A spectrum that contains a series of
discrete lines. Each line corresponds
to a specific wavelength.
A spectrum that contains continuous
band of light radiation of all
wavelengths.
Each line separated by blank area Continuous. No blank area in between.
Source: emission spectrum of atom
(from gas discharge tube of
hydrogen, etc.)
Source: white light, sunlight
2 They are two types of spectrum: line spectrum, and
continuous spectrum.
nucleus n=1
n=2
n=3
n=4
When the electron of a hydrogen
atom at its ground state absorbs
sufficient amount of energy, it will
move to a higher energy level
At higher energy evel, electron is
unstable. It will falls back to a lower
energy level.
During the transition (falls form
higher to lower energy level),
energy will be released in a form of
light at certain wavelength and
frequency.
Since energy is quantised (fix in
value), line spectrum is produced.
The electromagnetic spectrum: 3
There is several emission series of lines obtained during
the transition of electrons in hydrogen spectrum 4
The series in the hydrogen line spectrum are found in the
visible region, infrared region and ultraviolet region. 5
series Spectrum region
Lyman
Balmer
Paschen
Brackett
Pfund
Ultraviolet
Visible
Infrared
Infrared
Infrared
These spectral lines are the result of the transition of
excited electron to the lower energy level. 6
Balmer series is the only one visible to the unaided eye. It
contains four coloured line against a black background.
7
Line
number
Colour in
Balmer series
Wavelength
(nm)
1st line red 656
2nd line Blue-green 486
3rd line indigo 434
4th line violet 410
The possible energy levels in the hydrogen atom and the transitions
of electrons that produce the lines in the hydrogen spectrum: 7
Lines produced caused by the transition of electrons from
higher energy level to lower energy level. 7
Series
Line
Lyman
Transition of e-
1st
2nd
3rd
4th
5th
n=
n=1
n=2
n=3
n=4
n=5
n=6 n=7
n=2 to n=1
ch
hΔE
E , ,
Continuum
limit
e-
n=3 to n=1
e- n=4 to n=1
e-
n=5 to n=1
e-
n=6 to n=1
e-
Series
Line
Balmer
Transition of e-
1st
2nd
3rd
4th
n=
n=1
n=2
n=3
n=4
n=5
n=6 n=7
Continuum
limit
n=3 to n=2
e-
n=4 to n=2 e-
n=5 to n=2
e-
n=6 to n=2
e-
ch
hΔE
E , ,
Series
Line
Paschen
Transition of e-
1st
2nd
3rd
4th
n=
n=1
n=2
n=3
n=4
n=5
n=6 n=7
n=4 to n=3
Continuum
limit
e- n=5 to n=3
e-
n=6 to n=3
e- e-
n=7 to n=3
ch
hΔE
E , ,
Series
Line
Brackett
Transition of e-
1st
2nd
3rd
n=
n=1
n=2
n=3
n=4
n=5
n=6 n=7
n=5 to n=4
Continuum
limit
e-
n=6 to n=4
e-
n=7 to n=4
e-
ch
hΔE
E , ,
Series
Line
Pfund
Transition of e-
1st
2nd
n=
n=1
n=2
n=3
n=4
n=5
n=6 n=7
Continuum
limit
n=6 to n=5
e-
n=7 to n=5
e-
ch
hΔE
E , ,
Each line in the hydrogen spectrum has a specific
wavelength. 8
9 The wavelength can be calculated by using the
Rydberg equation:
2
2
2
1
1
11
nnRH
Where = wavelength
RH = Rydberg constant = 1.097 x 107 m-1
n1 and n2 = 1, 2, 3, …, (principal quantum numbers)
Since should have a positive value thus n1 < n2
nm441.3
m 10 x 4.41
m106 x 2.27
m 10 x 1.097
7-
1-
1-7
22 5
1
2
1
Example
21 n n ; 2
22
1
111
nnRH
Calculate the wavelength of the indigo line in the
Balmer series.
Solution:
Indigo line 3rd line in Balmer series n=5 to n=2
m nm 1 x 10-9
Try!!!
1. Calculate the wavelength of the continuum limit in the
Paschen series.
2. For the Lyman series, calculate the wavelength when
electron falls from n=4 to n=1.
• The further they fall, more energy is released and the higher the frequency.
• The orbitals also have different energies inside energy levels.
Ultraviolet Visible Infrared
hcΔE hvΔE
Changing the Energy
The following diagram depicts the line spectrum of hydrogen atom. Line A is the first line of the Lyman series.
47
Line spectrum
Specify the increasing order of the radiant energy, frequency and wavelength of the emitted photon.
Which of the line that corresponds to
i) the shortest wavelength?
ii) the lowest frequency?
A B C D E E
∆
Example
Line spectrum
A B C D E
Paschen series
Solution
Which of the line in the Paschen series corresponds to the longest wavelength of photon?
Describe the transition that gives rise to the line.
Example
Describe the transitions of electrons that lead to the lines W, and Y, respectively.
Solution
Line spectrum
W Y
Balmer series
Example
Compare the difference:
i f
E = RH 1
n2
1
n2
1 2
RH
1
n2
1
n2
1 =
n1 can be ni or nf as long as n1 < n2 to get positive value for
Both equations can be used to calculate wavelength ( ) and frequency ( ) of any line of H atom emission series
Remember! RH values and units are different in both equations !
RH = 2.18 x 10–18 J
RH = 1.097 x 107 m-1
Use the Rydberg equation to calculate the wavelength (in nm) of the forth line in the Balmer series of Hydrogen spectrum .
EXAMPLE
ANS: 410.2 nm
ANS:n = 3
An electron in the n = 5 level of an H atom emits a photon of wavelength 1281 nm. To what energy level does the electron move?
EXAMPLE
Calculate what is;
i ) Wavelength
ii ) Frequency
iii ) Wave number
of the last line of
hydrogen spectrum
in Lyman series
Wave number = 1/wavelength
For Lyman series; n1 = 1
& n2 = ∞
EXAMPLE
The minimum energy required to remove
one mole of electrons from one mole of
the gaseous atoms or ions.
M (g) → M+ (g) + e- ΔH1= +ve kJ
Calculate the ionization energy of H atom
in kJmol-1.
EXAMPLE
Ans: EXAMPLE
• Unable to explain the line spectrum of atoms or ions
containing more than one electron (such as helium).
• Electron is restricted to move in a certain distance around
the nucleus of an atom.
• Unable to explain the extra lines formed in the hydrogen
spectrum.
• Unable to explain the dual nature of electrons.
Bohr’s theory has a number of weaknesses. Bohr’s atomic
model
(A) de Broglie’s postulate
• in 1924, Louis de Broglie, a French physicist, suggested
that light and matter appear to have dual natures, that is,
both light and matter are wave-like as well as particle-
like.
• de Broglie suggested an equation that allows the
calculation of the wavelength of an electron or a particle
with mass (m), moving at velocity ( ):
m
h λ
where, h = Planck’s constant (J s)
m = particle mass (kg)
= velocity (m/s)
= wavelength of a matter wave
(B) Heisenberg’s uncertainty principle
• Bohr’s theory:
the electron was thought of as orbiting the nucleus.
This would mean that at any moment, we would
know both the precise position and precise
speed of the electron.
This is not allowed in quantum mechanics.
• In 1927, Werner Heisenberg states in his principle that:
This principle is known as Heisenberg’s uncertainty principle.
“It is impossible to know simultaneously both the
momentum, p (defined as mass times velocity) and the position of a particle with certain”.
where
x= uncertainty in measuring the position
p=uncertainty in measuring the momentum
= mv
h = Planck’s constant
• Mathematically, his principle stated as:
4
h px
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