39.2.
Model: Light of frequency f consists of discrete quanta, each of energy E  hf .
Solve: (a) The energy of the light quantum is
E
 hf
 h c



 
34



9


1 eV
1.6 10

19 J

3.11 eV
From Table 39.1, the work functions for sodium and potassium are smaller than 3.11 eV. That is, light of wavelength
400 nm has enough energy to eject photoelectrons from sodium and potassium.
(b) The energy of the light quantum is
E
 hf
 h c



 
34


9



1 eV
1.6 10

19 J

4.97 eV
Light of wavelength 250 nm has enough energy to eject photoelectrons from all of the metals on the table except gold.
39.4.
Solve: From Equation 39.7, the maximum kinetic energy is hc
E
0

K max


K max

 hf

E
0
 h c


E
0

34




1 eV
1.6 10

19
J
Assess:   209 nm is the wavelength of light in the ultraviolet region of the spectrum.

209 nm
39.5.
Model: The threshold frequency for the ejection of photoelectrons is
Solve: The visible region of light extends from 400 nm to 700 nm. For 
0 f
0
 E
0
/ h where E
0
is the work function.
 400 nm, the work function is
E
0
 f h
 hc

0


 
34


9



1 eV
1.6 10

19
J

3.11 eV
For 
0
 700 nm,
E
0


 
34

 8
6.63 10 J s 3.0 10 m/s


9

1 eV
1.6 10

19
J

1.78 eV
The cathode that will work in the entire visible range must have a work function of 1.78 eV or less.
39.8.
Solve: (a) The frequency of the photon is f

9


From Equation 39.4, the energy is
E
 hf


(b) The frequency of the photon is
 
15
 f

E
 h
5000 eV
4.14 10

15 eV s





1.77 eV
Thus, the wavelength is
  c
 f


2.5 10

10
m

0.25 nm
Assess: Because x-ray photons are very energetic, their wavelength is small.

39.13.
Model: All the light emitted by the light bulb is assumed to have a wavelength of 600 nm.
Solve: (a) The frequency of the emitted light is f
 
9

5.0 10 Hz
(b) The rate of photon emission is
R photon

P 5 W
= hf hf


 
34
5 W



   
39.20.
Model: To conserve energy, the emission and the absorption photons must have exactly the energy lost or gained by the atom in the appropriate quantum jumps.
Visualize: The energy of a light quantum is E  hf  hc/  .
Solve: (a) The wavelength of the emission photon from the n  2 to n  1 transition is
  hc
E
2

E
1


 
15

1.5 eV



828 nm
Likewise,   497 nm for the 3  2 transition with  E  2.5 eV, and   311 nm for the 3   transition with  E  4.0 eV.
(b) Because the atom in the ground state is in the n  1 state, the absorption lines correspond to the 1   and
1  3 transitions. The absorption wavelengths are 828 nm and 311 nm. The 2   transition is not seen in absorption.
39.21.
Model: To conserve energy, the absorption spectrum must have exactly the energy gained by the atom in the quantum jumps.
Visualize: Please refer to Figure EX39.20.
Solve: (a) An electron with a kinetic energy of 2.00 eV can collide with an atom in the n  1 state and raise its energy to the n  2 state. This is possible because E cannot be excited to the n  3 state.
2
– E
1
 1.50 eV is less than 2.00 eV. On the other hand, the atom
(b) The atom will absorb 1.50 eV of energy from the incoming electron, leaving the electron with 0.50 eV of kinetic energy.
39.23.
Model: The electron must have ground state.
Visualize: k
 
E atom to cause collisional excitation. The atom is initially in the n  1
Solve: The kinetic energy of the incoming electron is
E

1
2 mv
2 
1
2

 
31

 6

2   
19
J

4.81 eV
The electron has enough energy to excite the atom to the n  2 stationary state ( E
2
– E
1
 4.00 eV). However, it does not have enough energy to excite the atom into the n  3 state which requires a total energy of 6.00 eV.
39.34.
Solve: The laser light delivers 2.50  10 17 photons per second and 100  10  3 J of energy per second. Thus, the energy of each photon is



3

1

4.00 10

19 J
From Equation 39.4, the wavelength of the photons is
  c
 hc
 f E

 
34

4.00 10

19 J



4.97 10 m

497 nm
Assess: The wavelength is in the visible region.
39.51.
Model: Photons are emitted when an atom undergoes a quantum jump from a higher energy level to a lower energy level. On the other hand, photons are absorbed in a quantum jump from a lower energy level to a higher energy level. Because most of the atoms are in the n

1 ground state, the only quantum jumps in the absorption spectrum start from the n  1 state.
Solve: (a) The ionization energy is
E
1

6.5 eV.
(b) The absorption spectrum consists of the transitions 1  2 and 1  3 from the ground state to excited states.
According to the Bohr model, the required photon frequency and wavelength are f


E h
   c
 hc f

E where  E  E f
– E i
is the energy change of the atom. Using the energies given in the figure, we calculated the values in the table below.
Transition E f
(eV) E i
(eV)  E (eV)  (nm)
1  2
1  3
 3.0
–2.0


6.5
6.5
3.5
4.5
355
276
(c) Both wavelengths are ultraviolet (  < 400 nm).
(d) A photon with wavelength match  E of the atom, a 1240 nm photon can be emitted only in a 3  2 transition. So, after the collision the atom was
  1240 nm has an energy in
E photon
 hf  hc /   1.0 eV. Because E n  3 state. Before the collision, the atom was in its ground state ( n  1). Thus, an electron with v photon
must exactly i
 1.4  10 the
 J of kinetic energy. Initially, the kinetic energy of the electron was
6 m/s collided with the atom in the n  1 state. The atom gained 4.5 eV in the collision as it is was excited from the n  1 to n
 3, so the electron lost 4.5 eV  7.20  
K i

1
2 m v 2

1
2

 
31


6

2   
19 J
After losing 7.20  10  19 J in the collision, the kinetic energy is
K f

K i

7.20 10

19
J
  
19
J

1
2 m v
2  
2 K f m elec



19 J
9.11 10

31 kg

  5
6.16 10 m/s