Walker4 ISM Ch31

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Chapter 31: Atomic Physics

Answers to Even-Numbered Conceptual Questions 2. There are many such reasons, but perhaps the most important is that orbiting electrons in Rutherfords

model would radiate energy in the form of electromagnetic waves, with the result that atoms would collapse in a very small amount of time.

4. The observation that alpha particles are sometimes reversed in direction when they strike a thin gold foil led to the idea that there must be a great concentration of positive charge and mass within an atom. This became the nucleus in Rutherfords model.

6. In principle, there are an infinite number of spectral lines in any given series. The lines become more closely spaced as one moves higher in the series, which makes them hard to distinguish in practice.

8. (a) There is no upper limit to the wavelength of lines in the spectrum of hydrogen. The reason is that the wavelength is inversely proportional to the energy difference between successive energy levels. The spacing between these levels goes to zero as one moves to higher levels, and therefore the corresponding wavelengths go to infinity. (b) There is a lower limit to the wavelength, however, because there is an upper limit of 13.6 eV to the energy difference between any two energy levels.

10. All of these questions can be answered by referring to Figure 31-17 and Table 31-3. (a) Not allowed; there is no d subshell in the n = 2 shell. (b) Not allowed for two reasons. First, there is no p subshell in the n = 1 shell. Second, a p subshell cannot hold seven electrons. (c) Allowed. (d) Not allowed; the n = 4 shell does not have a g subshell.

12. No. Atoms in their ground states can emit no radiation. Even if an electron dropped from a highly excited state to the ground state in one of these atoms, the result would not be an X-ray. The reason is that the binding energy of these atoms is simply much lower than the energy of a typical X-ray photon.

Solutions to Problems and Conceptual Exercises 1. Picture the Problem: In this problem we are given the radius of the nucleus and radius of the atom and want to

calculate the fraction of the volume occupied by the nucleus.

Strategy: To find the fraction of the volume occupied by the nucleus, divide the volume of the nucleus by the volume of the atom. Assume the nucleus and atoms are both spheres.

Solution: 1. Write the ratio of volumes in terms of the radius:

334nnucleus 3 n

34atom aa3

rV rV rr

= =

2. Insert the given radii:

31516nucleus

11atom

0.50 10 m 8.4 105.3 10 m

VV

= =

Insight: Most of the mass of the atom is contained in the nucleus, but it accounts for a tiny portion of the total volume.

Chapter 31: Atomic Physics James S. Walker, Physics, 4th Edition


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2. Picture the Problem: A model of an atom is constructed by using a baseball to represent the nucleus.

Strategy: In the previous problem we were told that the diameter of a hydrogen nucleus is 151.0 10 m and that the electron is typically found about 115.3 10 m from the nucleus. Use these values to calculate the ratio of the diameter of the nucleus to the radius of the atom, and then use the ratio to find the model electrons distance from the center of the baseball.

Solution: 1. Set the ratio of the nucleus to atom equal to the ratio of the baseball to the new electron distance and solve for the electron distance:

nucleus baseball baseballelectron atom

atom electron nucleus

d d d

r rr r d

= =

2. Insert the given distances: ( )2 11electron 157.3 10 m 5.3 10 m 3.9 km1.00 10 mr = =

Insight: If the nucleus were the size of a baseball, the atom would be the size of a small town. 3. Picture the Problem: An alpha particle of charge 2e is brought from infinity to the surface of a copper nucleus.

Strategy: Let the initial configuration correspond to the alpha particle at rest and infinitely far from the nucleus, and let the final configuration correspond to the alpha particle at rest at a distance of one nuclear radius. The work required to bring the alpha particle near the nucleus is the nonconservative work nc f iW E E= (equation 8-9), but because the kinetic energy is zero in both configurations, nc .W U= Use equation 20-8 to solve for the change in potential energy.

Solution: 1. (a) Find the work required to bring an alpha particle from infinity to a distance r:

( )( ) 20nc

2 29 580k e ekq q keW U

r r r = = = =

2. Insert the constants and set the distance equal to half the diameter of the nucleus:

( )( )( )

29 2 2 19

1512

58 8.99 10 N m C 1.60 10 C5.6 pJ

4.8 10 mW

= = Insight: The energy may seem small, but it is equivalent to 35 MeV, over 68 times the rest energy of an electron! 4. Picture the Problem: The image shows an alpha particle

approaching a gold nucleus with an initial kinetic energy of 3.0 MeV. The alpha particle comes to rest at a distance d from the gold nucleus when all of the kinetic energy has been converted to electric potential energy.

Strategy: Set the initial kinetic energy equal to the potential energy at the point d, using equation 20-8. Solve the resulting equation for the distance d.

Solution: 1. Set the initial kinetic energy equal to the final potential energy:

kqQKd

=

2. Solve for the separation distance:

2(2 )(79 ) 158kqQ k e e kedK K K

= = =

3. Insert the given data and constants:

( )( )( )( )

29 2 2 19

6 19

158 8.99 10 N m C 1.60 10 C76 fm

3.0 10 eV 1.60 10 J/eVd

= =

Insight: Because the alpha particle moves away from the nucleus without colliding, this experiment verifies that the gold nucleus has a radius smaller than 76 fm.



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5. Picture the Problem: The image shows several of the Balmer series spectral lines. We

want to calculate the wavelength of the line corresponding to n = 17.

Strategy: Use equation 31-1 to calculate the wavelength, setting n = 17.

Solution: 1. Solve equation 31-1 for the inverse of the wavelength: 2 2

1 1 12

Rn

=

2. Insert the numerical values: ( )7 1 7 12 21 1 11.097 10 m 0.2705 10 m2 17 = =

3. Invert the result to obtain : 77 11 3.697 10 m 369.7 nm0.2705 10 m

= = =

Insight: This wavelength is in the ultraviolet part of the spectrum. 6. Picture the Problem: The Balmer series corresponds to transitions in atomic hydrogen that terminate at n = 2.

Strategy: We want to find the smallest value of n for which the wavelength is less than 400 nm. Set the wavelength in equation 31-1 equal to 400 nm and solve for the value of n. Because n must be an integer, round the solution up to the next integer.

Solution: 1. Set the wavelength equation to 400 nm in eq. 31-1: 2 2

1 1 1 1400 nm 2

Rn

= =

2. Solve for the value of n: ( ) ( )( )2 2 9 7 1

1 1 1 1 1 0.0221400 nm 42 400 10 m 1.097 10 m

1 6.70.0221

Rn

n

= = =

= =

3. Round n to the next highest integer: = 7n

Insight: As a check that we have the correct answer, note that the Balmer wavelength for n = 6 is 410 nm, and is 397 nm for n = 7. Thus n = 7 is the smallest value of n for which the wavelength is less than 400 nm.

7. Picture the Problem: The image shows the energy levels for the three longest wavelength

transitions in the Lyman series.

Strategy: Use equation 31-2 to calculate the appropriate wavelengths. For the Lyman series set n = 1. The longest wavelengths correspond to the smallest values of n. Therefore, set n equal to 2, 3, and 4 to obtain the longest wavelengths.

Solution: 1. Set n = 1 in eq. 31-2 and solve for the wavelength:

2

111Rn

=

2. Set n = 2: ( )( )2 7 1 2

1 121.5 nm1.097 10 m 1 1 2

= =

3. Set n = 3: ( )( )3 7 1 2

1 102.6 nm1.097 10 m 1 1 3

= =

4. Set n = 4: ( )( )4 7 1 2

1 97.23 nm1.097 10 m 1 1 4

= =

Insight: All three of these wavelengths lie in the ultraviolet portion of the electromagnetic spectrum.



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8. Picture the Problem: The image shows the energy levels for the three longest wavelength

transitions in the Paschen series.

Strategy: Use equation 31-2 to calculate the appropriate wavelengths. For the Paschen series set n = 3. The longest wavelengths correspond to the smallest values of n. Therefore, set n equal to 4, 5, and 6 to obtain the longest wavelengths.

Solution: 1. Set n = 3 in eq. 31-2 and solve for the wavelength:

2 2

11 13

Rn

=

2. Set n = 4: ( )4 7 1 2 2

1 1875 nm1 11.097 10 m3 4

= =

3. Set n = 5: ( )5 7 1 2 2

1 1282 nm1 11.097 10 m3 5

= =

4. Set n = 6: ( )6 7 1 2 2

1 1094 nm1 11.097 10 m3 6

= =

Insight: All three of these wavelengths lie in the infrared portion of the electromagnetic spectrum. 9. Picture the Problem: The image shows two electron transitions in hydrogen. The transition

at lower right is a member of the Lyman series, and the transition at upper left is in the Paschen series. We want to calculate the longest wavelength in the Lyman series and the shortest wavelength in the Paschen series.

Strategy: The longest wavelength in a series has the smallest change in energy. In the Lyman series this corresponds to the transition n = 2 to n = 1. The shortest wavelength in a series has the largest energy transition. In the Paschen series this transition corresponds to n = to n = 3. Use equation 31-2 to calculate the corresponding wavelengths.

Solution: 1. (a) Calculate the longest wavelength in the Lyman series: ( )2 22 1 11 2

1 121.5 nmR

= =

2. (b) Calculate the shortest wavelength in the Paschen series: ( )213

1 820.4 nm0R

= =

Insight: The longest wavelength in the Lyman series is ultraviolet while the shortest in the Paschen is infrared. Visible light is therefore not possible from either series.

10. Picture the Problem: The figure shows the transitions the lead to the longest and shortest

wavelengths in the Brackett (n = 4), Paschen (n = 3), and Balmer (n = 2) series. We want to know if the Brackett and Paschen series overlap and if the Paschen and Balmer series overlap.

Strategy: For each of the series calculate the longest and shortest wavelengths using equation 31-2. The longest wavelength occurs for n = n + 1. The shortest wavelength occurs for n = . Compare the ranges between series to determine if they overlap.

Solution: 1. (a) Solve eq. 31-2 for the wavelength: ( )

17 -1

2 2

1 11.097 10 mn n

=



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2. Calculate the shortest wavelength in the Brackett series (n = 4, n = ): ( )

17 -1

2

1 11.097 10 m 1459 nm4

= =

3. Calculate the longest wavelength in the Brackett series (n = 4, n = 5): ( )

17 -1

2 2

1 11.097 10 m 4051 nm4 5

= =

4. Write Brackett series range: Brackett1459 nm < 4051 nm < 5. Calculate the shortest wavelength

in the Paschen series (n = 3, n = ): ( )1

7 -12

1 11.097 10 m 820.4 nm3

= =

6. Calculate the shortest wavelength in the Paschen series (n = 3, n = 4): ( )

17 -1

2 2

1 11.097 10 m 1875 nm3 4

= =

7. Write Paschen series range, and compare with the Brackett series:

Paschen820.4 nm < 1875 nm < The Brackett and Paschen series overlap.

8 (b). Calculate the shortest wavelength in the Balmer series (n = 2, n = ): ( )

17 -1

2

1 11.097 10 m 364.6 nm2

= =

9. Calculate the shortest wavelength in the Balmer series (n = 2, n = 3): ( )

17 -1

2 2

1 11.097 10 m 656.3 nm2 3

= =

10. Write Balmer series range, and compare with the Paschen series:

Balmer364.6 nm < 656.3 nm < The Paschen and Balmer series do not overlap.

Insight: The Lyman and Balmer series are the only ones that do not overlap with any other series. 11. Picture the Problem: Suppose the mass of the electron were magically doubled.

Strategy: The ionization energy of hydrogen depends upon the energy of the n = 1 orbit. Equation 31-8 indicates that the magnitude of that energy is linearly proportional to the mass of the electron. This result follows from equation 31-5, where we see that the radius of an electrons orbit is inversely proportional to the mass, and from equation 31-6, where we see that the speed of the electron is independent of its mass. Use these relationships to answer the question.

Solution: 1. (a) Referring to equation 31-8, we can see that the total (negative) energy of the electron would double with its mass, as would the energy required to ionize the atom. We conclude that if the mass of the electron were magically doubled, the ionization energy of hydrogen would increase.

2. (b) The best explanation is I. The ionization energy would increase because the increased mass would mean the electron would orbit closer to the nucleus and would require more energy to move to infinity. Statements II and III are each false.

Insight: Statement III is false because the force holding the electron in orbit is the electrical force, not the gravitational force, so the mass does not cancel out of the equations.



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12. Picture the Problem: Three atoms or ions have a single electron orbiting the nucleus: (A) neutral hydrogen in the state

n = 2; (B) singly ionized helium in the state n = 1; and (C) doubly ionized lithium in the state n = 3. Strategy: The Bohr model can be used to describe atoms or ions that have a single electron. Use the expression

22

2 24nhr nmk Ze

= (equation 31-7) to determine the ranking of the Bohr radii.

Solution: Noting that the radius of Bohr orbits is proportional to 2 ,n Z we find that for atom (A), n = 2 and Z = 1, so 2 22 1 4.n Z = = For atom (B) 2 21 2 1 2,n Z = = and for atom (C) 2 23 3 3.n Z = = We conclude that B < C < A.

Insight: For each energy state n the orbit radii get progressively smaller as Z increases because the electron is more tightly bound to the nucleus as the amount of positive charge in the nucleus increases.

13. Picture the Problem: Three atoms or ions have a single electron orbiting the nucleus: (A) neutral hydrogen in the state

n = 3; (B) singly ionized helium in the state n = 2; and (C) doubly ionized lithium in the state n = 1. Strategy: The Bohr model can be used to describe atoms or ions that have a single electron. Use the expression

( ) 2213.6 eVn ZE n= (equation 31-9) to determine the ranking of the Bohr energies. Solution: Noting that the radius of Bohr orbits is proportional to ( )2 2 ,Z n we find that for atom (A), n = 3 and

Z = 1, so ( ) ( )2 21 3 1 9.Z n = = For atom (B) ( ) ( )2 22 2 1,Z n = = and for atom (C) ( ) ( )2 23 1 9.Z n = = Recall that a small negative energy is greater than a large negative energy, so C < B < A.

Insight: For each energy state n the energy becomes progressively more negative as Z increases because the electron is more tightly bound to the larger amount of positive charge in the nucleus.

14. Picture the Problem: An electron in the n = 1 Bohr orbit has the kinetic energy K1.

Strategy: Use the expressions in the derivation leading to equation 31-8 to note that 2

.2

keZE Kr

= = Use this relationship and the fact that 21nE n to determine the relationship between K2 and K1.

Solution: We can see from the derivation leading up to equation 318 that the kinetic energy of a Bohr orbit is equal to minus one times the total energy of the orbit; that is, .K E= In addition, recall that E is proportional to 21 .n Therefore, the kinetic energy of an electron in the n = 2 Bohr orbit is 12 14 .K K=

Insight: We can also find the answer by means of a ratio: ( )( )

22 222 2 1

2 12 21 1 1 2

13.6 eV 1 1 1 2 4 413.6 eV

nK E n K KK E n n

= = = = = = 15. Picture the Problem: The Bohr model of the hydrogen atom can be used to predict the speed of an electron in any

quantum state. Strategy: The speed of the electron can be written in terms of known constants using equation 31-6. Set n = 3 and

divide the result by the speed of light.

Solution: 1. Divide eq. 31-6 by the speed of light:

22nv kec nhc

=

2. Set n = 2 and insert known constants: ( )( )( )( )( )29 2 2 19

328 34

2 8.99 10 N m /C 1.60 10 C2.42 10

3 3.00 10 m/s 6.63 10 J svc

= = Insight: The speed is much less than the speed of light, so that the effects of relativity on the energy and the momentum

can be neglected.



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16. Picture the Problem: The Bohr model of the hydrogen atom can be used to calculate the force between the proton and

electron.

Strategy: Use Coulombs Law (equation 19-5) to calculate the attractive force between the proton and electron. The separation distance is given by the radius of the first Bohr orbit, 111 5.29 10 mr

= (equation 31-5).

Solution: Solve Coulombs Law for the force on the electron:

( )( )( )

29 2 2 1928

2 211i

8.99 10 N m / C 1.60 10 C8.22 10 N

5.29 10 m

keFr

= = =

Insight: This force creates a centripetal acceleration of 22 29.02 10 m/s for the electron, which keeps it in orbit around the proton.

17. Picture the Problem: An amount of energy equal to the binding energy of the electron must be added to a hydrogen

atom in order to ionize it.

Strategy: The Bohr model can be used to determine the binding energy of an electron in any quantum state. The energy necessary is equal to the energy difference between n = and the n = 5 Bohr orbits. Use equation 31-9 to calculate the binding energy:

Solution: Use equation 31-9 to find bindingE : binding 4 2

13.6 eV0 0.544 eV5

E E E = = =

Insight: Ionizing a hydrogen atom from the ground state requires 13.6 eV, significantly more than that required to ionize from the n = 5 state. However, the excitation of the atom to the n = 5 state from the ground state requires 13.6 eV 0.544 eV = 13.1 eV.

18. Picture the Problem: The energy of the electron is determined by its Bohr orbit. To switch between orbits, a photon

with energy equal to the difference in orbital energies must be absorbed by the electron.

Strategy: Set the energy of the absorbed photon equal to the change in electron energies between the n = 2 and n = 6 orbits. The orbit energies are given by equation 31-9.

Solution: Use equation 31-9 to find :E ( )5 2 2 21 113.6 eV 3.02 eV6 2E E E

= = = Insight: This energy corresponds to a photon of wavelength 410 nm, in the violet portion of the visible spectrum. 19. Picture the Problem: An electron in the nth Bohr orbit has the potential energy Un.

Strategy: Use the expressions in the derivation leading to equation 31-8 to note that

2

2 .2

keZE Ur

= = Use this relationship and the fact that 21nE n to determine the relationship between Un and Un+1.

Solution: We can see from the derivation leading up to equation 318 that the potential energy of a Bohr orbit is equal to two times the total energy of the orbit; that is, 2 .U E= In addition, recall that E is proportional to 21 .n Therefore,

the potential energy of an electron in the n +1 Bohr orbit is 2

1 .1n nnU U

n+ = +

Insight: We can also find the answer by means of a ratio: ( ) ( )

( ) ( )2 22

1 112 2

13.6 eV 12

2 113.6 eV 1n n

n nn n

nU E n nU UU E nn n

+ ++

+ = = = = = + +



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20. Picture the Problem: The Bohr model of the hydrogen atom can be used to calculate the linear momentum and the

angular momentum for an electron in the n = 3 state.

Strategy: Calculate the linear momentum of the electron by multiplying the electron mass by its velocity in the n = 3 state, where the velocity is given by equation 31-6. Calculate the angular momentum by multiplying the linear momentum by the orbital radius, which is given by equation 31-5.

Solution: 1. (a) Multiply the mass by the orbital velocity:

22n n

kep mv mnh

= =

2. Set n = 3 and solve for the linear momentum:

( )( )( )( )

231 9 2 2 19

3 34

25

2 9.11 10 kg 8.99 10 N m C 1.60 10 C

3 6.63 10 J s

6.62 10 kg m/s

p

= =

3. (b) Multiply the orbital radius by the momentum:

2 2 2

2 2

224n n n

h n mke hnL r pnhmke

= = =

4. Set n = 3 and solve for the angular momentum:

3434

33 3(6.63 10 J s) 3.17 10 J s2 2

hL = = =

Insight: Note that the linear momentum varies inversely with n, but the angular momentum increases linearly with n. 21. Picture the Problem: The Bohr model can be used to calculate the kinetic energy, potential energy, and total energy of

an electron in the n = 3 state of the hydrogen atom.

Strategy: Calculate the kinetic energy of the electron, 212 ,K mv= where the velocity in the n = 3 state is given by equation 31-6. Calculate the potential energy, 2 ,U ke r= using the orbital radius given by equation 31-5. Sum the potential and kinetic energies to calculate the total energy.

Solution: 1. (a) Find the speed of the electron:

( )( )( )( )

29 2 2 192

3 34

5

2 8.99 10 N m /C 1.60 10 C23 6.63 10 J s

7.270 10 m/s

kevnh

= = =

2. Calculate the kinetic energy: ( )( )22 31 51 13 2 219

19

9.11 10 kg 7.270 10 m/s

1 eV2.408 10 J 1.51 eV1.602 10 J

K mv

= = = =

3. (b) Write the potential energy in terms of the Bohr radius:

2 2

21

nn

ke keUr n r

= =

4. Calculate the potential energy for n = 3:

( )( )( )

29 2 2 19

3 2 11

1919

8.99 10 N m / C 1.60 10 C

3 5.29 10 m

1 eV1.89 10 J 3.02 eV1.602 10 J

U =

= =

3. (c) Sum the kinetic and potential energies: 3 3 3 1.51 eV 3.02eV 1.51 eVE K U= + = = Insight: Note that the total energy is the negative of the kinetic energy. If you write the kinetic energy and potential

energy in terms of fundamental constants, you will find that 2n nU K= and n nE K= .



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22. Picture the Problem: When energy is added to an electron in the n = 3 orbit, it will jump to a higher orbit. We want to

calculate the final energy level attained when 1.23 eV is added to such an electron.

Strategy: Calculate the energy in the n = 3 orbit using equation 31-9 (with Z = 1). Add the 1.23 eV to the result to calculate the energy of the final orbit. Then use equation 31-9 to find the value of n in the final state of the electron.

Solution: 1. Calculate the energy of the n = 3 orbit: 3 2

13.6 eV 1.51 eV3

E = =

2. Calculate the final energy level: 3 1.51 eV 1.23 eV = 0.281 eVnE E E= + = +

3. Solve for the quantum number n: 2

13.6 eV 13.6 eV 13.6 eV 70.281 eVn n

E nEn

= = = =

Insight: If the electron were to now spontaneously drop back down to its ground state, it would give off an ultraviolet photon that has energy ( )0.281 eV 13.6 eV 13.3 eVE = = and wavelength 93.1 nm.

23. Picture the Problem: The Bohn model can be used to find the values of n for the initial and final states of a hydrogen

atom when the wavelength of the emitted photon is known.

Strategy: Use equation 31-2 to relate the emitted wavelength to the initial and final states. Only one combination of states will correspond to the emitted wavelength. We could find these states by trial and error. However, because a wavelength of 656 nm falls within the visible spectrum, and only the Balmer series overlaps the visible spectrum, we know that nf = 2. Insert nf = 2 into equation 31-2 and solve for ni.

Solution: 1. Solve equation 31-2 for ni:

12

i2 2 2f i f

1 1 1 1 1 R nRn n n

= =

2. Insert the wavelength and nf = 2: ( )( )

12

i 2 9 7 1

1 1 32 656 10 m 1.097 10 m

n

= =

3. Write the transition: The transition is i f3 2n n= = Insight: This problem could have been solved by trying different values for the initial and final states. However, we

simplified the problem by noting the wavelength ranges of the different series in the hydrogen spectrum. 24. Picture the Problem: An electron in a hydrogen atom absorbs a photon and jumps from the n = 3 to the n = 5 state

according to the Bohr model of the atom. Another electron absorbs a photon and jumps from the n = 5 to the n = 7 state.

Strategy: We want to calculate the energy of the photon that will cause these transitions. The energy of the absorbed photon must be equal to the difference in the energies of the two electron states. Calculate the difference in energy levels using equation 31-9 (with Z = 1).

Solution: 1. (a) Calculate change in energy: f i 2 2 2 2

f i i f

13.6 eV 13.6 eV 1 113.6 eVE E En n n n

= = =

2. Set ni = 3 and nf = 5: 35 2 2

1 113.6 eV 0.967 eV3 5

E = =

3. (b) The energy of the photon would be less than that found in part (a) because the energy levels that correspond to higher values of n are spaced more closely together (see Figure 31-8) so that the energy difference must be smaller.

4. (c) Set ni = 5 and nf = 7: 57 2 2

1 113.6 eV 0.266 eV5 7

E = =

Insight: The energy difference between levels rapidly decreases as n increases (see equation 31-9 and Figure 31-8).



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25. Picture the Problem: The problem lists four transitions in a hydrogen atom. The Bohr model can be used to find the

transitions that correspond to the emission or absorption of photons with the largest and shortest wavelengths.

Strategy: Use equation 31-2 to calculate the wavelength associated with each transition. From the resulting wavelengths determine the transition with the longest wavelength and the transition with the shortest wavelength. The atom will lose energy any time the electron falls to an orbit of smaller n.

Solution: 1. Solve equation 31-2 for the wavelength: ( )

1

7 -12 2

1 11.097 10 mf in n

=

2. Set ni = 2, nf = 6: ( ) 17 -1( i ) 2 21 11.097 10 m 410.2 nm6 2

= =

3. Set ni = 2, nf = 8: ( ) 17 -1( ii ) 2 21 11.097 10 m 388.9 nm8 2

= =

4. Set ni = 7, nf = 8: ( ) 17 -1( iii ) 2 21 11.097 10 m 19.06 m8 7

= =

5. Set ni = 6, nf = 2: ( ) 17 -1( iv ) 2 21 11.097 10 m 410.2 nm2 6

= =

6. (a) Transition (iii) i f7 8n n= = has the longest wavelength of 19.06 m. 7. (b) Transition (ii) i f2 8n n= = has the shortest wavelength of 388.9 nm. 8. (c) Transition (iv) i f6 2n n= = is the only one in which n decreases and the atom loses energy. Insight: Instead of calculating each wavelength, we could have determined the shortest and longest wavelengths from

the energy levels. Transition (iii) is between two large sequential levels, so it would correspond to the longest wavelength. Transitions (i), (ii), and (iv) all start or end in the n = 2 level. Because transition (ii) has the largest change in the quantum number n, it will correspond to the shortest wavelength.

26. Picture the Problem: The electron in a hydrogen atom is replaced by a muon, which has the same charge, but has a

mass that is 207 times heavier than the electron. The Bohr theory can be used to calculate how the increased mass will affect the orbital radius of the electron and the wavelengths of light emitted in the Balmer series.

Strategy: Replace the mass of the electron in equation 31-5 with the mass of a muon to calculate the radius of muonium. In equation 31-10 we see that the constant R is proportional to the mass of the electron. Therefore, the Rydberg constant for muonium is R = 207 R. Use this fact to solve equation 31-10 for the longest wavelength in the Balmer series of muonium. The longest wavelength corresponds to the transition 3 2n n= = .

Solution: 1. (a) Calculate the radius of muonium: ( )

( )

2 2

12 2 2 2

11 13

12074 4 207

1 5.29 10 m 2.56 10 m207

e

h hr rm ke m ke

= = =

= =

2. (b) The wavelengths in the Balmer series of muonium will be less than those for hydrogen because the wavelength is inversely proportional to the Rydberg constant R, which is proportional to the particles mass.

3. (c) Calculate the longest wavelength: ( )

( )2 2

17 -1

2 2

1 1 1207

1 1207 1.097 10 m 3.17 nm2 3

Rn n

= = =

Insight: The longest wavelength in the Balmer series of hydrogen is 656 nm, which is in the visible spectrum. The Balmer series of muonium lies in the X-ray region of the electromagnetic spectrum.



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27. Picture the Problem: Doubly ionized lithium has a single electron orbiting a nucleus of charge +3e. We want to

calculate the radius of the n = 4 orbit and the energy difference between the n = 4 and n = 5 energy levels.

Strategy: Use equation 31-7 to write the radius in terms of the Bohr radius and the nuclear charge, Z. Calculate the difference in energy levels for hydrogen by using equation 31-9 with Z = 1, and for Li2+ by using Z = 3.

Solution: 1. (a) Write equation 31-7 in terms of r1:

22 2 2 21

2 2 2 24 4nn rn h n hr

Z ZmkZe mke = = =

2. Set n = 4 and Z = 3: ( )11 10

4

16 5.29 10 m2.83 10 m

3r

= =

3. (b) The energy required to raise an electron from the n = 4 state to the n = 5 state in 2Li + is greater than that for hydrogen because the force on the electron due to the three protons in the 2Li + nucleus is stronger than that due to the single proton in the hydrogen nucleus.

4. (c) Calculate the energy difference for hydrogen: ( )

( )( )f

2H 2 2

i f

22 2

1 113.6 eV

1 113.6 eV 1 0.306 eV4 5

in nE E E Z

n n = =

= =

5. Calculate the energy difference for Li2+: ( ) ( )22Li H 3 0.306 eV 2.75 eVE Z E = = = Insight: The energy levels for lithium are Z2 = 9 times larger in magnitude than the levels for hydrogen because the

orbits are smaller by a factor of Z and the charge of the nucleus is larger by a factor of Z. Inserting the two Zs into the equation for total energy results in a multiplicative factor of Z2 (see equation 31-8).

28. Picture the Problem: A triply ionized beryllium atom has a nucleus of charge +4e that is orbited by a single electron.

The Bohr model can be used to calculate the shortest wavelength in the Lyman series and the binding energy for Be3+.

Strategy: The wavelength of light emitted from an electron transition is inversely proportional to the difference between the initial and final energy levels (equation 31-10). From equation 31-9 we see that the energy levels of a hydrogen-like atom are equal to the energy levels in the hydrogen atom multiplied by the square of the nuclear charge Z.To find the wavelengths of the spectrum of Be3+, multiply the Rydberg constant in equation 31-2 by Z2 = 42. The shortest wavelength in the Lyman series (n = 1) occurs when n , so insert these values for n and n. The ionization energy is the absolute value of the ground state energy. Use equation 31-9 to calculate the ionization energy.

Solution: 1. (a) Multiply the Rydberg constant in equation 31-2 by Z2 and solve for the wavelength:

12

2 2

1 1RZn n

=

2. Insert n = 1 and n = and calculate : ( )( ) 17 -1 2 1 11.097 10 m 4 5.697 nm

1

= =

3. (b) Set n = 1 in equation 31-9: ( ) ( )2 213.6 eV 4 13.6 eV 218 eVE Z = = = Insight: The energy needed to remove the last electron from a Be3+ ion is 16 times greater than the energy required to

remove the electron from a hydrogen atom.



31 12

29. Picture the Problem: An electron in the n = 2 orbit of a hydrogen atom remains in this orbit for 108 s before dropping

down to the n = 1 orbit. The Bohr model can be used to find the time required to complete one orbit and the number of orbits completed before dropping to the lower orbit.

Strategy: The time required to complete one orbit is the circumference of the orbit divided by the speed. Write the circumference as 2 times the radius that is given by equation 31-5. Divide the circumference by the velocity that is given in equation 31-6 to calculate the time. Calculate the number of orbits by dividing the lifetime of the orbit by the time to complete one orbit.

Solution: 1. (a) Set the time for one orbit equal to the circumference divided by the velocity, where vn is given by equation 31-6:

2 22 2 4

2

n n n n

n n

n

C r r mrt

v v nhnhmr

= = = =

2. Write the radius in terms of the Bohr orbit and insert the constants:

( )( )( ) ( )

22 2 2 2 31 1

2 32 31 11

34

15

4 4

4 9.11 10 kg 5.29 10 m 2

6.626 10 J s1.22 10 s 1.22 fs

m r n mr ntnh h

t

= = =

= =

3. (b) Divide the orbit lifetime by the period of one orbit:

86

15

10 s 8 10 orbits1.22 10 s

=

Insight: Even though the lifetime of the orbit is only 10 ns, the electron is able to complete about 8 million revolutions in that time.

30. Picture the Problem: The Bohr model of the hydrogen atom can be used to correlate the kinetic energy of the electron

with the quantum number n.

Strategy: Write the kinetic energy in terms of the electron mass and orbital speed (given by equation 31-6) and solve for the quantum number n. Repeat for the next higher orbit by replacing n with n + 1 in the kinetic energy equation.

Solution: 1. (a) Write the kinetic energy using equation 31-6 for vn:

2221 1 2

2 2n nkeK mv m

nh = =

2. Solve for the orbit number:

( )( ) ( )

2

29 2 2 19 31

34 19

2

8.99 10 N m C 1.60 10 C 2 9.11 10 kg4

6.63 10 J s 1.35 10 J

n

ke mnh K

=

= =

3. (b) The kinetic energy will decrease as the electron moves to a higher orbit because K is inversely proportional to n.

4. (c) Replace n with n + 1: ( )

2 22 22 221

1 121 2 1 22 1 1 2 1n n n

ke n ke nK mv m m Kn h n nh n

+ + = = = = + + +

5. Set n = 4 to solve for the kinetic energy of the n + 1 = 5 orbit: ( )

219 19

5 42

4 16 1.35 10 J 0.864 10 J 0.544 eV255

K K = = = =

Insight: Because the kinetic energy is always the negative of the total energy in a Bohr orbit, the orbit number could

have equivalently been solved using the absolute value of equation 31-9: 19

19

13.6 eV 1.602 10 J 4.1 eV1.35 10 J

n

= =



31 13

31. Picture the Problem: The Bohr model of the hydrogen atom can be used to correlate the potential energy of the

electron with the quantum number n.

Strategy: Write the electrostatic potential energy of the atom (equation 20-8) in terms of the Bohr radius. Solve the resulting equation for the orbit number. To calculate the potential of the next higher orbit, replace n by n + 1 in the potential energy equation.

Solution: 1. (a) Insert the Bohr radius into equation 20-8:

2 2

21

nn

ke keUr n r

= =

2. Solve for the orbit number: ( ) ( )( )( )

29 2 2 192

11 191

8.99 10 N m / C 1.60 10 C6

5.29 10 m 1.20 10 Jn

kenrU

= = =

3. (b) If n increases, the potential energy will become a smaller negative number and thus increase. This is because positive work is done on the electron to move it farther away from the nucleus.

4. (c) Set n = 7 in the potential energy equation:

( )( )( )

29 2 2 192

7 2 117

19

8.99 10 N m / C 1.60 10 C

7 5.29 10 m

0.888 10 J 0.555 eV

keUr

= = = =

Insight: The potential energy is equal to twice the total energy, so the orbit number could also have been found by using equation 31-9.

32. Picture the Problem: As two hydrogen atoms collide head-on, their kinetic energies are converted into the internal

energy required to boost each electron into the n = 3 quantum state.

Strategy: If the two atoms have the same speed but are traveling in opposite directions before the collision, their net momentum is zero. The entire kinetic energy of each atom in the collision can be used to excite that atom to the n = 3 state. Use equation 31-9 to calculate the difference in energies between the ground (n = 1) and second excited state (n = 3). Set the kinetic energy of the atom equal to this energy difference and solve for the speed of the atom.

Solution: 1. Solve equation 31-9 for the energy necessary to excite the atom: 2 1 2 2

1918

1 113.6 eV3 1

1.60 10 J12.1 eV 1.94 10 JeV

E E E

= = = =

2. Set the change in energy equal to the kinetic energy and solve for the speed: ( )

212

184

27

2 1.94 10 J2 4.81 10 m/s 48.1 km/s1.674 10 kg

K mv E

Evm

= = = = = =

Insight: This speed is much greater than the room temperature rms speed of 1.92 km/s for atoms in a hydrogen (H2) gas. For this reason, almost all hydrogen atoms are typically found in their ground state at room temperature.



31 14

33. Picture the Problem: When an electron jumps from the n to the n 1 orbit in a hydrogen atom, it emits a photon of a

specific frequency. The electron also orbits the nucleus in a Bohr orbit with given frequency.

Strategy: The photon frequency is the speed of light divided by the photon wavelength. Use equation 31-10 to write the wavelength in terms of the known constants and variables. The electron orbit frequency is the electron speed divided by the circumference of the orbit. Use the electron speed given by equation 31-6 and the Bohr orbit given by equation 31-5 to calculate the electron orbit frequency. The n-dependence of the frequency of the emitted photon consists of the difference between two fractions. Combine these fractions and take the limit for large n. Substitute this back into the photon frequency and compare with the orbital frequency.

Solution: 1. (a) Use equation 31-10 to write the frequency emitted as an electron jumps to the next lower orbit (n 1): ( )

2 2 4

photon 3 2 2

1 2 1 11

mk ef ch nn

= =

2. (b) Set the orbit frequency equal to the orbit speed divided by the circumference:

2

2 2 4

electron 3 32 2

2 2

21 4

22

4

n

n

kenhv mk ef

T r n hn hmke

= = = =

3. (c) Write the n-dependence of the photon frequency as a single fraction: ( )

( )( ) ( )

22 2 2

2 2 2 2 22

4 3 2

11 1 2 12 11 1

2 12

n n n n nn n n nn n n

nn n n

+ = = + = +

4. Take the limit as n becomes large:

4 3 2 3

2 1 2lim2nn

n n n n = +

5. Substitute the limit into the photon frequency equation:

2 2 4 2 2 4

photon electron3 3 3 3

2 2 4limn

mk e mk ef fh n n h

= = =

For very large n, the frequency of the emitted photon is the same as that of the electrons orbital motion.

Insight: For small values of n, the orbital frequency is greater than the frequency of the emitted photon. 34. Picture the Problem: The de Broglie wavelengths of electrons in two different Bohr orbits are compared.

Strategy: Note that one de Broglie wavelength fits around the circumference of the n = 1 orbit, but two de Broglie wavelengths fit around the circumference of the n = 2 orbit.

Solution: 1. (a) The de Broglie wavelength of an electron in the n = 2 Bohr orbit is greater than the wavelength of an electron in the n = 1 orbit. First, the n = 2 Bohr orbit has a circumference that is 4 times greater than that of the n = 1 orbit. Second, two de Broglie wavelengths fit into the n = 2 orbit, compared to one in the n = 1 orbit. Combining these results, we see that the de Broglie wavelength in the n = 2 orbit is twice as long as that in the n = 1 orbit.

2. (b) The best explanation is I. The de Broglie wavelength in the nth state is 2r/n, where r is proportional to n2. Therefore, the wavelength increases with increasing n, and is greater for n = 2 than for n = 1. Statement I ignores the increase in r with increasing n, and statement III ignores the fact that the de Broglie wavelength also depends on v.

Insight: In a similar fashion, the de Broglie wavelength of an n = 4 electron is twice as long as the de Broglie wavelength for an n = 2 electron.



31 15

35. Picture the Problem: The Bohr model can be used to calculate the de Broglie wavelength of an electron in the ground

state of a hydrogen atom.

Strategy: The de Broglie wavelength is given in equation 30-16 in terms of the electron momentum. Write the momentum as the product of mass and velocity, and substitute the velocity as given by equation 31-4.

Solution: 1. Write the de Broglie wavelength in terms of the velocity in the nth orbit: n n nh p h mv = =

2. Write the velocity using equation 31-4: ( )

22

nn

n

rhm nh mr n

= =

3. Set n = 1: ( )1111 2 2 5.29 10 m 0.332 nm1r = = = Insight: Equation 31-5 indicates that the radius is proportional to the square of n. So in general, 1n n = . 36. Picture the Problem: An electron in the nth state of a hydrogen atom has a corresponding de Broglie wavelength.

Strategy: The de Broglie wavelength is given in equation 30-16 in terms of the electron momentum. Write the momentum as the product of mass and velocity, with the velocity given by equation 31-6.

Solution: 1. Write the de Broglie wavelength in terms of the velocity in the nth state: n

n n

h hp mv

= =

2. Write the velocity using equation 31-6: ( )2

22 22nh nh

mkem ke nh = =

Insight: In general, the de Broglie wavelength is n times the ground state wavelength, or ( )1 0.332 nm .n n n = = 37. Picture the Problem: The image shows the de Broglie wavelength associated with a certain

Bohr orbit for a hydrogen atom.

Strategy: There are five wavelengths contained in the orbit shown in the figure, so we conclude the electron is in the n = 5 state. Use equation 31-5 to calculate the radius of the n = 5 orbit.

Solution: Calculate the radius for n = 5:

( )2

1

2 115 5 5.29 10 m 1.32 nmnr n r

r

== =

Insight: The de Broglie wavelength is proportional to the orbit number, and the number of wavelengths contained in the circumference is equal to the orbit number. These two facts require the radius to be proportional to n2.

38. Picture the Problem: An electron that has a de Broglie wavelength equal to the size of an atom (51011 m = 0.5 ) or

the size of a nucleus (1015 m = 1 fm) has a definite momentum and a corresponding kinetic energy.

Strategy: Write the kinetic energy as 2 2K p m= and use equation 30-16 to write the momentum in terms of the de Broglie wavelength.

Solution: 1. (a) Solve for the kinetic energy in terms of the de Broglie wavelength:

22 2

2

12 2 2p h hKm m m

= = =

2. Insert a wavelength of 0.5 : ( )( )( )234

2 1931 10

6.63 10 J s 1 eV 0.6 keV1.60 10 J2 9.11 10 kg 0.5 10 m

K = =

3. (b) Insert a wavelength of 1015 m:

( )( )( )

23412

2 1931 15

6.63 10 J s 1 eV 2 10 eV1.60 10 J2 9.11 10 kg 10 m

K = =

Insight: Note that the energies on a nuclear scale are much greater than those on the atomic scale.



31 16

39. Picture the Problem: The quantum mechanical model of the atom restricts the values of A that are allowed when n =

5.

Strategy: Equation 31-11 states that the allowed orbital angular momentum quantum numbers are: 0, 1, 2, , ( 1)n= A . Insert n = 5 and write out the possible quantum numbers.

Solution: Set n = 5 in equation 31-11: 0, 1, 2, , (5 1) 0, 1, 2, 3, 4= =A Insight: There is always the same number of allowed orbital angular momentum quantum states as the principal

quantum number. 40. Picture the Problem: The quantum mechanical model of the atom restricts the values of mA that are allowed for n = 4.

Strategy: Equation 31-13 gives the allowed values for the magnetic quantum number: , 1, 2, , 1, 0, 1, , 2, 1, m = + + A A A A A A A . Equation 31-11 states that the maximum value of A is one less

than n. Find the maximum value of A and use it to determine the possible values of .mA Solution: 1. Calculate the maximum value of A : max 1 4 1 3n= = =A 2. Write out the possible values of .mA 3, 2, 1, 0, 1, 2, 3m =A 3. Count how many states are possible: There are 7 possible values of .mA

Insight: In general, there are ( )2 1n possible values of mA for any given value of n. 41. Picture the Problem: The quantum mechanical model of the atom relates the orbital angular momentum L to the

quantum number A . Strategy: Equation 31-12 gives the magnitude of the orbital angular momentum in terms of A . Insert the given values

into this equation for L and calculate the integer A , if it exists. Solution: 1. (a) Solve equation 31-12 for A : ( ) ( ) ( )

( )1 / 2 6 / 2

1 6 2

L h h = + =+ = =

A AA A A

2. (b) Repeat for ( )15 2 :L h = ( ) ( ) ( )( )

1 / 2 15 / 2

1 15 no solution exists

L h h = + =+ =

A AA A

3. (c) Repeat for ( )30 2 :L h = ( ) ( ) ( )( )

1 / 2 30 / 2

1 30 5

L h h = + =+ = =

A AA A A

4. (d) Repeat for ( )36 2 :L h = ( ) ( ) ( )( )

1 / 2 36 / 2

1 36 no solution exists

L h h = + =+ =

A AA A

Insight: Not all values of L are possible because A is limited to integer values. 42. Picture the Problem: Hydrogen atom 1 is in the 4f state and hydrogen atom 2 is in the 5d state. The energy and orbital

angular momentum of each are described by the quantum mechanical model of the atom.

Strategy: Use equation 31-9 to calculate the energy in the 4f state, where n = 4. Use equation 31-12 to calculate the orbital angular momentum, where 3=A corresponds to the subshell f. To compare the energies of atoms 1 and 2, compare their principal quantum numbers. To compare the orbital angular momenta, compare their subshells.

Solution: 1. (a) Set n = 4 in equation 31-9:

( )4 2

13.6 eV0.850 eV

4E = =



31 17

2. (b) Set 3=A in equation 31-12: ( ) ( ) ( )34 343 4 6.63 10 J s1 3.66 10 J s

2 2hL

= + = = A A

3. (c) The energy in atom 2 is greater than the energy in atom 1, because an n = 5 state is farther from the nucleus and has a smaller negative energy.

4. (d) The orbital angular momentum in atom 1 is greater than the orbital angular momentum in atom 2, because 3=A for an f sublevel and 2=A for a d sublevel.

Insight: The principle quantum number (or shell number) determines the energy of the electron and the subshell determines the orbital angular momentum.

43. Picture the Problem: The quantum mechanical model of the atom relates the orbital angular momentum to the quantum number A and the minimum possible n.

Strategy: Use equation 31-12 to solve for the orbital quantum number A and then use equation 31-11 to find the minimum value of the principal quantum number n. Use equation 31-9 and n to determine the energy of the atom.

Solution: 1. (a) Insert the given value of L into equation 31-12 and solve for the orbital quantum number:

( ) ( ) ( )( ) ( )

1 / 2 10 57 / 2

1 5700 75 76

75

h h + =+ = =

=

A AA A

A

2. (b) Use equation 31-11 to find the minimum value of n: max max1 1 75 1 76n n= = + = + =A A

3. (c) Calculate the energy for n = 76: ( )76 213.6 eV 0.00235 eV 2.35 meV

76E = = =

Insight: To find the value of the integer A , where ( )1 ,X+ =A A it is simplest to take the square root of X. The result lies between A and 1+A because ( ) ( )22 1 1< + < +A A A A . Find the integer immediately below X and multiply it by the integer above to verify that the product is X. An alternative approach is to apply the quadratic formula to find A .

44. Picture the Problem: The quantum mechanical model of the atom relates the energy E and the maximum orbital angular momentum L to the quantum numbers s, , , and .n m mAA

Strategy: Use equation 31-9 and the given energy to calculate the principal quantum number. For a given value of A there are 2 1+A possible values of mA , and for each value of mA there are 2 possible values of s .m Thus, there are

( )2 2 1+A states in each A sublevel. Set the number of states equal to 18 and solve for .A Insert the orbital quantum number into equation 31-12 to calculate the orbital angular momentum. Finally, subtract one from A and calculate the number of possible states.

Solution: 1. (a) Solve equation 31-9 for n:

2

13.6 eV 13.6 eV 50.544 eVn

E nn

= = =

2. (b) Find the A that corresponds to 18 states: ( ) 1 1818 2 2 1 1 4

2 2 = + = = A A

3. Insert 4=A into equation 31-12: ( ) ( )34 344 5 6.63 10 J s 4.72 10 J s

2L

= =

4. (c) Calculate the number of states for 3:=A ( ) ( )# of states = 2 2 1 2 2 3 1 14+ = + =A Insight: Note that the orbital angular quantum number A is less than the principal quantum number n, as required by

equation 31-11.



31 18

45. Picture the Problem: The quantum mechanical model of the atom relates the orbital angular momentum L to the

quantum numbers and .mAA Strategy: According to equation 31-13, the maximum magnetic quantum number is equal to the orbital angular

momentum quantum number, ,maxm =A A . Use the given values for the maximum magnetic quantum number to calculate the ratio of orbital angular momenta from equation 31-12.

Solution: 1. (a) Because the maximum magnetic quantum number in state I is greater than that in state II, IA is greater than IIA and, therefore, IL is greater than II .L

2. (b) Calculate the ratio of orbital angular momenta, using I ,I 3m= =AA and II ,II 2m= =AA :

( ) ( )( ) ( )

( )( )

I II

II II II

1 2 3 3 12

1 2 2 2 1

hLL h

+ += = =+ +A AA A

Insight: Use the expression below equation 31-13 to verify for yourself that the ratio Z,I Z,II 3 2.L L = 46. Picture the Problem: The quantum mechanical model of the atom stipulates the number of electrons that can occupy

each subshell and shell.

Strategy: Use the Pauli exclusion principle together with the quantum mechanical model of the atom to determine the number of electrons that can occupy a given subshell.

Solution: 1. (a) The maximum number of electrons allowed in any given subshell depends only on the quantum number( ), , ,... ,s p dA and not on the quantum number ( )K,L, M,... .n In the case of a p subshell, mA can take on 3 values:

0, 1.m = A For each of these three values, ms can take on two values. Thus, 6 electrons can occupy the 2p subshell. 2. (b) Using the same reasoning as in part (a), 6 electrons can occupy the 3p subshell.

Insight: In a similar manner, any d subshell can hold a maximum of 10 electrons, regardless of the value of n. 47. Picture the Problem: The quantum mechanical model of the atom stipulates the number of electrons that can occupy


Strategy: Use the Pauli exclusion principle together with the quantum mechanical model of the atom to determine the number of electrons that can occupy a given subshell.

Solution: 1. (a) The maximum number of electrons allowed in any given subshell depends only on the quantum number( ), , ,... ,s p dA and not on the quantum number ( )K,L, M,... .n In the case of a d subshell, mA can take on 5 values:

0, 1, 2.m = A For each of these five values, ms can take on two values. Thus, 10 electrons can occupy the 3d subshell.

2. (b) The n = 2 shell has an s subshell that can accommodate 2 electrons and a p subshell that can accommodate 6 electrons, for a total of 8 electrons that can occupy the n = 2 shell.

Insight: In a similar manner, any f subshell can hold a maximum of 14 electrons, regardless of the value of n. 48. Picture the Problem: The quantum mechanical model of the atom stipulates the number of electrons that can occupy


Strategy: Add the exponents in the given configuration to determine the total number of electrons in the atom.

Solution: Adding the numbers in the exponents, we find that this atom has 2 2 6 2 1 13+ + + + = electrons. Insight: The configuration is a shorthand notation that makes it easy to indicate the arrangement of all electrons in an

atom.



31 19

49. Picture the Problem: The figure shows how the levels are filled for a carbon atom.

Strategy: A carbon atom has six protons (Z = 6) and six electrons. Start filling at the n = 1 shell, for which there is only an s subshell. The 1s-subshell is filled with two electrons. Then start filling the n = 2 shell. First fill the 2s subshell with two electrons, and then place the remaining two electrons in the 2p subshell.

Solution: Write out the configuration for carbon: 2 2 21 2 2s s p

Insight: Table 31-4 gives the electronic configurations for the first 19 elements. 50. Picture the Problem: Each electron in a neon atom has a unique set of the four quantum numbers.

Strategy: Neon is a noble gas with ten electrons. The ten electrons will completely fill the n = 1 and n = 2 shells. Create a table with a row for each electron. Start with the n = 1 shell. For this shell 0=A , 0,m =A and 12sm = . In the n = 2 shell, fill the 2s subshell first with 0=A , 0,m =A and 12sm = . Then fill the 2p subshell with 1=A , 1,0,1m = A , and

12sm = .

Solution: 1. Fill in the 1s shell with two electrons:

electron n A mA sm

11s 1 0 0 12 21s 1 0 0 12

12s 2 0 0 12 22s 2 0 0 12

12 p 2 1 1 12 22 p 2 1 1 12

32 p 2 1 0 12 42 p 2 1 0 12

52 p 2 1 1 12 62 p 2 1 1 12

2. Fill in the 2s shell with two electrons:

3. Fill in the 2p shell with six electrons:

Insight: Because the n = 2 shell is completely filled, neon is an inert gas. 51. Picture the Problem: The figure shows how the levels are filled for nitrogen.

Strategy: Nitrogen has seven protons (Z = 7) and seven electrons. Fill the 1s shell with the first two electrons. Fill the 2s subshell with the next two electrons. Place the last three electrons in the 2p subshell.

Solution: Write the electronic configuration for nitrogen: 2 2 31 2 2s s p

Insight: The electronic configurations for the first 19 elements are given in Table 31-4.



31 20

52. Picture the Problem: Each electron in a 3s subshell has a unique set of the four quantum numbers.

Strategy: The 3s subshell has n = 3 and 0.=A Since 0,=A 0.m =A The quantum number sm can take on values of 12 . Write out the two possible sets of quantum numbers in tabular format.

Solution: Fill the 3s subshell with two electrons:

n A mA sm

3 0 0 12 3 0 0 12

Insight: The number of electrons in any s subshell is always 2, regardless of the principal quantum number. 53. Picture the Problem: Each electron in a 3p subshell has a unique set of the four quantum numbers.

Strategy: In the 3p subshell, n = 3, and 1.=A Because 1,=A mA can take on the values 1, 0, and 1, and sm can take on the values of 12 . Write out the six possible sets of quantum numbers in tabular format.

Solution: Write all six different combinations for mA and sm , where n = 3 and 1:=A

n A mA sm

3 1 1 12 3 1 1 12

3 1 0 12 3 1 0 12

3 1 1 12 3 1 1 12

Insight: The number of electrons that can occupy p subshell is always 6, regardless of the principal quantum number.



31 21

54. Picture the Problem: Each electron in a magnesium atom has a unique set of the four quantum numbers.

Strategy: Magnesium has twelve electrons arranged in the configuration 1s2 2s2 2p6 3s2 (Table 31-4). Create a table with a row for each electron. Start with the n = 1 shell. For this shell 0=A , 0,m =A and 12sm = . In the n = 2 shell, fill the 2s subshell first with 0=A , 0,m =A and 12sm = . Then fill the 2p subshell with six electrons with quantum numbers 1=A , 1,0,1m = A , and 12sm = . Finally, fill the 2s subshell with the last two electrons with quantum numbers 0=A , 0,m =A and 12sm = .

Solution: 1. Fill the 1s shell with 2 electrons:

n A mA sm 11s 1 0 0 12 21s 1 0 0 12

12s 2 0 0 12 22s 2 0 0 12

12 p 2 1 1 12 22 p 2 1 1 12

32 p 2 1 0 12 42 p 2 1 0 12

52 p 2 1 1 12 62 p 2 1 1 12

13s 3 0 0 12 23s 3 0 0 12

2. Fill the 2s subshell with 2 electrons:

3. Fill the 2p subshell with 6 electrons:

4. Fill the 3s subshell with the last 2 electrons:

Insight: If the atom were to have any additional electrons, they would begin to fill the 3p subshell. 55. Picture the Problem: The configuration of the outer 10 electrons for nickel is given.

Strategy: Nickel has 28 protons (Z = 28) and 28 electrons. The configuration of the outer 10 electrons is given as 8 23 4 .d s Start filling the subshells beginning at 1s. Continue with the 2s, 2p, 3s, and 2p subshells until you have listed

18 electrons. Add the remaining 10 electrons in the given configuration. Each s subshell is filled with 2 electrons and the p subshells are filled with six electrons.

Solution: 1. Write out the electronic configuration for the first 18 electrons: 2 2 6 2 61 2 2 3 3s s p s p

2. Add the last ten electrons as specified: 2 2 6 2 6 8 21 2 2 3 3 3 4s s p s p d s

Insight: Note that the 4s subshell fills before the 3d shell is filled, because the 4s subshell has a lower energy (see Figure 31-15).



31 22

56. Picture the Problem: Each electron in a shell n has a unique set of the four quantum numbers.

Strategy: For each ,A there are (2 1)+A values for mA and two values for .sm Therefore for each A there will be 2 (2 1)+A states possible. For each n, the values of A range from 0 to n 1. For n = 2, sum the number of states possible for A = 0 and A = 1. For n = 3, sum the states from n = 2 with the added states for A = 2. For n = 4, sum the states from n = 3 with the added states for A = 3.

Solution: 1. (a) Sum the possible states for A = 0 and A = 1: ( ) ( ) ( )total states 2 2 2 0 1 2 2 1 1 2 6 8n = = + + + = + =

2. (b) Add the possible states for A = 2: ( ) ( ) ( )total states 3 total states 2 2 2 2 1 8 10 18n n= = = + + = + = 3. (c) Add the possible states for A = 3: ( ) ( ) ( )total states 4 total states 3 2 2 3 1 18 14 32n n= = = + + = + = Insight: For each higher value of n, the number of possible states will increase because of the additional value of A . 57. Picture the Problem: Each electron in a shell n has a unique set of the four quantum numbers.

Strategy: The number of states in any subshell is equal to 2 (2 1)+A . In any shell the value of A ranges from 0 to n 1. Write the number of states as the sum of the states in each subshell for the entire range of subshells in each shell. Write out this summation and note that the summation is related to the square of the principal quantum number.

Solution: 1. Write the total number of states in a shell as the sum of the number of states in each subshell: ( ) ( ) ( )

1 1

0 0total states 2 2 1 2 2 1

n n

n

= == + = +

A AA A

2. Write out the summation in detail: ( ) ( ){ }( )

total states 2 1 3 5 2 1 1

2 1 3 5 2 1

n n

n

= + + + + + = + + + +

3. Note that the term in the brackets is equal to n2: ( )2 1 3 5 2 1n n= + + + + 4. Simplify the expression from step 2: ( ) 2total states 2n n= Insight: You can use this formula to verify the numbers of states found in Problem 56: ( ) 2total states 2 2 2 8= = ,

( ) 2total states 3 2 3 18,= = and ( ) 2total states 4 2 4 32.= =



31 23

58. Picture the Problem: Each of the 10 electrons in a 5d subshell has a unique set of the four quantum numbers.

Strategy: The 5d notation refers to the subshell for which n = 5 and 2.=A The fact that 2=A means that the magnetic quantum numbers mA range from 2 to +2. For each mA there are two states corresponding to 1s 2 .m = Use these facts to write out the ten possible states in tabular format.

Solution: Write out the states starting with the lowest value of :mA

n A mA sm 15d 5 2 2 12 25d

5 2 2 12

35d

5 2 1 12

45d

5 2 1 12

55d

5 2 0 12

65d

5 2 0 12

75d

5 2 1 12

85d

5 2 1 12

95d

5 2 2 12

105d

5 2 2 12

Insight: Note that each set of quantum numbers is unique and that the 10 quantum numbers include all possible combinations in the 5d subshell.

59. Picture the Problem: X-rays are produced when high-energy electrons bombard a metal target and remove an inner-

shell electron from the target atom. An X-ray photon is emitted when an outer electron jumps down into the vacancy.

Strategy: Note the principles involved in the production of X-rays to answer the conceptual question.

Solution: 1. (a) The wavelength of characteristic X-rays depends only on the type of atom in the target, and not on the energy of the incoming electrons. Therefore, the wavelength of the characteristic X-rays will stay the same.

2. (b) The best explanation is III. The wavelength of characteristic X-rays depends only on the material used in the metal target, and does not change if the energy of incoming electrons is increased. Statements I and II are each false.

Insight: Increasing the energy of the incoming electrons will shift the peak of the Bremsstrahlung continuum to shorter wavelengths, but the characteristic peaks will remain at the same wavelengths.



31 24

60. Picture the Problem: A high-energy photon is absorbed by a fluorescent material that later emits a low-energy photon.

Strategy: Use the relationship between energy and wavelength of a photon to answer the conceptual question.

Solution: The radiation that excites a fluorescent material is less than the wavelength of the radiation it emits (see Figure 31-29, for example). Note that a smaller wavelength implies a higher frequency, and hence a higher energy for the corresponding photon. This, in turn, is in agreement with the physical mechanism illustrated in Figure 31-28.

Insight: Fluorescent lamps use a coating to convert the ultraviolet 254-nm emission from a Hg-Ar discharge into the white visible light that we see.

61. Picture the Problem: The image shows an electron in the n = 2 shell of nickel

dropping down to the n = 1 shell and producing a K photon.

Strategy: To find the wavelength of the K photon, start with the relationship between the change in energy of an electron and the wavelength of the corresponding photon, E hf hc = = . Use equation 31-14 to find the energy of an electron in the K shell of nickel using Z = 28. Because the nucleus is partially screened from the electron in the L shell by the single electron in the K shell, calculate the energy of the electron in the L shell using equation 31-14 again, except with n = 2. Use the energy difference to calculate the wavelength.

Solution: 1. Calculate the change in energy between the n = 2 and n = 1 levels in nickel: ( ) ( ) ( )

2 2

K L 2 2

28 1 28 113.6 eV 7.44 keV

1 2E E E

= = = 2. Solve the photon energy equation

for , noting that hc = 1240 eVnm: 1240 eV nm 0.167 nm

7440 eVhc hcE

E

= = = =

Insight: This wavelength falls in the X-ray portion of the electromagnetic spectrum. 62. Picture the Problem: The image shows an electron in the n = 2 shell of lead

dropping down to the n = 1 shell and producing a K photon.

Strategy: Use equation 31-14 to find the energy of an electron in the K shell of lead using Z = 82. Because the nucleus is partially screened from the electron in the L shell by the single electron in the K shell, calculate the energy of the electron in the L shell using equation 31-14 again, except with n = 2. The difference between the two energy levels is equal to the energy of the emitted K photon.

Solution: Calculate the change in energy between the n = 2 and n = 1 levels in nickel: ( ) ( ) ( )

2 2

K L 2 2

82 1 82 113.6 eV 66.9 keV

1 2E E E

= = = Insight: The energy of the K photon increases as the square of the atomic number. Therefore, the energy of the K

photon for lead (66.9 keV) is much greater than the energy for the K photon of nickel (7.44 keV) found in Problem 61. 63. Picture the Problem: The image shows an electron in the L shell of iron dropping

down to the K shell and producing a K photon.

Strategy: The energy difference between the K and L shells is equal to the difference in ionization energies of the two shells. Solve for the wavelength using the relationship E hf hc = = and noting that hc = 1240 eVnm.

Solution: Solve for : 1240 eV nm 0.195 nm

8500 eV 2125 eVhcE

= = = Insight: Because iron has a smaller atomic number than nickel, the wavelength of the K photon for iron (0.195 nm) is

longer than the K of nickel (0.167 nm) found in Problem 61.



31 25

64. Picture the Problem: The image shows an electron in the L shell dropping down to

the K shell producing a K photon with a wavelength of 0.0205 nm.

Strategy: Calculate the energy between the two shells using the relationship E hf hc = = , and noting that hc = 1240 eVnm. Because the nucleus is

partially shielded from the electron in the L shell by the single electron in the K shell, the change in energy of the dropping electron can be found using equation 31-9, where Z is replaced with Z 1, as in equation 31-14. Set the change in energy equal to the energy of the photon and solve for Z.

Solution: 1. Calculate the energy of the photon from the wavelength:

1240 eV nm 60.5 keV0.0205 nm

hcE = = =

2. Set the energy equal to the change in energy of the electron: ( )( )

set2L K 2 2

1 113.6 eV 1 60.5 keV2 1

E E E Z = = =

3. Solve for Z: ( ) ( )2

34

60500 eV1 593013.6 eV

1 77 78

Z

Z Z

= = = =

Insight: The result Z = 78 indicates that the element is platinum. 65. Picture the Problem: Electrons are accelerated through a potential difference. They then hit a platinum target where

some electrons collide with K shell electrons, knocking them out of the atom. In some atoms, an electron from the L shell drops down to the K shell to fill the vacancy created by the missing electron, and a K photon is emitted.

Strategy: In order to produce a K X-ray, an electron in the K shell would have to be ejected from the atom by the collision with the moving electron. The moving electron then must have an energy at least equal to the binding energy of the K-shell electron. Use equation 31-14 to calculate the binding energy. To calculate the electric potential, divide the energy by the charge of the electron (e).

Solution: 1. (a) Set the minimum kinetic energy equal to the binding energy of the K-shell electron:

( ) ( )2

2min K 2

113.6 eV 13.6 eV 78 1 80.6 keV

1Z

K E= = = =

2. (b) Set the kinetic energy equal to the change in potential energy in the accelerator and solve for the potential:

minmin

80.6 keV 80.6 kVKq V K V

q e = = = =

Insight: In addition to K photons, this process will also produce K and longer wavelength photons. 66. Picture the Problem: The image shows laser pulses directed toward the cornea during

photorefractive keratectomy. The wavelength of the light is 193 nm.

Strategy: Calculate the energy difference between the two levels by setting the energy difference equal to the energy of the photon using the relationship ,E hf hc = = noting that hc = 1240 eVnm. Divide the total energy by the energy of each photon to calculate the number of photons needed.

Solution: 1. (a) Calculate the energy of the photon from the wavelength:

1240 eV nm 6.44 eV193 nm

hcE = = =

2. (b) Set the total energy equal to the energy per photon times the number of photons and solve for the number of photons:

total photon

135total

18photon

1.88 10 J 1.83 10 photons1.029 10 J

E n E

En

E

== = =

Insight: The photons from the laser interact with the molecules of the cornea, heating them rapidly and vaporizing them off the surface. The procedure changes the shape of the cornea to correct the persons vision.



31 26

67. Picture the Problem: A hydrogen atom makes the following three transitions: (A) i f5, 2;n n= = (B) i f7, 2n n= = ;

(C) i f7, 6n n= = . Strategy: The energy of the photon that is released when an atom makes a transition from a higher level orbit to a lower

level orbit is equal to the energy difference between the two states. Use the Bohr model to predict the energies of the photons released by the given transitions and then use E = hf (equation 30-4) and f = c (equation 25-4) to rank the frequencies and wavelengths of the photons.

Solution: 1. (a) Using ( ) 2 2

f i

1 113.6 eVEn n

= (equation 31-10) we can find the energies of the photons are EA =

2.86 eV, EB = 3.12 eV, and EC = 0.100 eV. Because is inversely proportional to f and therefore inversely proportional to E, we conclude that the ranking of the wavelengths is B < A < C.

2. (b) Because f is proportional to E, the ranking of the frequencies is C < A < B.

Insight: The larger the jump between energy levels, the bluer the emitted photon. 68. Picture the Problem: An electron is in the ground state orbit of hydrogen.

Strategy: The energy of the ground level orbit of hydrogen is 13.6 eV (equation 31-9). Use this fact to predict the highest energy photon that could be absorbed without dissociating the electron from the proton, and then find the energy required to make the smallest jump possible when a photon is absorbed by a hydrogen atom in its ground state. Such a transition corresponds to ni = 1 to ni = 2. Use equation 31-10 to find the energy of such a photon.

Solution: 1. (a) The highest possible energy level that does not dissociate the hydrogen atom corresponds to n = , or an energy of zero. The energy of the ground level orbit of hydrogen is 13.6 eV (equation 31-9), so the highest energy photon this system can absorb is 13.6 eV.

2. (b) Using ( ) 2 2

f i

1 113.6 eVEn n

= (equation 31-10) we can find that the lowest energy photon that a hydrogen

atom in its ground level orbit can absorb is ( ) 2 21 113.6 eV 10.2 eV .1 2E = =

Insight: These two photons are each in the ultraviolet portion of the electromagnetic spectrum. The 10.2 eV photon has a wavelength of 121.6 nm and the 13.6 eV photon has a wavelength of 91.2 nm.

69. Picture the Problem: The electronic configuration of a particular carbon atom is 2 2 1 11 2 2 3 .s s p s

Strategy: Compare this configuration with the configuration of the ground level of carbon presented in Table 31-4. Solution: The electronic configuration of a carbon atom in its ground energy level is 2 2 21 2 2s s p (Table 31-4). We can

see that the carbon atom presented in this question is in an excited state, because there is room for more electrons in the 2p subshell, but instead, this electron is in the 3s subshell. Because the energy of the 3s subshell is greater than that of the 2p subshell, this atom has more energy than it would have in its ground state.

Insight: The energies of various shells and subshells are depicted in Figure 31-18, which can be used to predict the lowest energy (ground) level configuration for most of the atoms in the periodic table.

70. Picture the Problem: The electronic configuration of a particular potassium atom is 2 2 6 2 6 11 2 2 3 3 3 .s s p s p d

Strategy: Compare this configuration with the configuration of the ground level of potassium presented in Table 31-4. Solution: The electronic configuration of a potassium atom in its ground energy level is 2 2 6 2 6 11 2 2 3 3 4s s p s p s

(Table 31-4). We can see that the potassium atom presented in this question is in an excited state, because the 3d1 orbit of the outermost electron has a higher energy than the 4s1 orbit of the ground level (see Figure 31-18).




31 27

71. Picture the Problem: The ionization energy of an atom is the minimum energy required to dissociate an electron from

the atom and leave the atom as a positively charged ion. Strategy: Compare the energies of the outermost electrons of sodium and lithium. The electron with the lower (more

negative) energy would be more difficult to remove from the atom and require a higher ionization energy. Solution: The outermost electron in sodium is in a higher energy level than the outermost electron of lithium.

Therefore, we expect the ionization energy of sodium to be less than that of lithium.


72. Picture the Problem: A photon with sufficient energy will ionize a hydrogen atom in its ground state.

Strategy: A photon incident on a hydrogen atom will ionize the atom if the energy of the photon is greater than the ionization energy of the atom. To find the minimum frequency, set the energy of the photon (equation 30-4) equal to the binding energy of hydrogen (equation 31-9) and solve for the frequency.

Solution: 1. Set photon binding :E E= 1 13.6 eVE hf E= = =

2. Solve for the frequency: ( )19 1534

13.6 eV 1.60 10 J/eV3.28 10 Hz

6.63 10 J sf

= =

Insight: This wavelength falls in the ultraviolet portion of the electromagnetic spectrum. 73. Picture the Problem: The temperature of atoms in a gas is related to the kinetic energy of the gas molecules. If the

temperature of hydrogen gas is high enough, the average kinetic energy would be sufficient to excite the atoms to the n = 2 excited state during a collision.

Strategy: Use equation 31-9 to calculate the energy required to excite an electron to the first excited state by calculating the difference in energies for the ground and excited states. Set this energy equal to the average kinetic energy of the hydrogen gas (equation 17-12) and solve for the temperature.

Solution: 1. Calculate the energy difference between the n = 1 state and the n = 2 state:

19 18

2 2

13.6 eV 13.6 eV 1.60 10 J10.2 eV 1.63 10 J1 eV1 2

E

= = =

2. Set avE K = and solve for T:

( )( )

18 av3

av 2 23

2 1.63 10 J2 78,700 K

3 3 1.38 10 J/KKK kT Tk

= = = =

Insight: This temperature is 13.6 times hotter than the temperature of the surface of the Sun (5800 K)!



31 28

74. Picture the Problem: The image shows a hydrogen atom initially at rest

with an electron in the n = 4 shell. The electron drops to the n = 2 shell emitting a photon in the positive x-direction. The atom recoils and gains momentum in the negative x-direction.

Strategy: Set the energy of the photon equal to the change in energy of the electron as it jumps from the n = 4 to the n = 2 orbit. Calculate the change in energy using equation 31-9. Use equation 30-11 to write the momentum of the photon. Set the momentum of the photon equal to the momentum of the atom and solve for the speed of the atom.

Solution: 1. (a) Set the energy of the photon equal to the change in energy of the electron: photon 2 2

19 19

13.6 eV 13.6 eV4 2

1.602 10 J2.55 eV 4.085 10 J1 eV

E

= = =

2. Calculate the momentum of the photon:

1927

8

4.085 10 J 1.36 10 kg m/s3.00 10 m/s

Epc

= = =

3. (b) Set photon atomp p= and solve for the recoil speed of the atom:

27

27

1.364 10 kg m/s 0.815 m/s1.674 10 kg

pvm

= = =

Insight: Because the photon has momentum, the atom must acquire an equal and opposite momentum for the net momentum to remain zero. The recoil speed is very small, however, compared to rms 1930 m/sv = for hydrogen molecules at room temperature.

75. Picture the Problem: The figure shows an electron jumping from the

n = 4 shell to the n = 2 shell of

Walker4 ISM Ch31

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