Irvington High School  AP Chemistry
Mr. Markic
Name _______________________________
Number ___ Date ___/___/___
7  Electronic Structure of Atoms
The Photoelectric Effect
1.
A photon has a wavelength of 624 nm. Calculate the energy of the photon in joules.
E  h 
2.
hc
(6.63  1034 J  s)(3.00  108 m/s)

 3.19  1019 J

624  109 m
The blue color of the sky results from the scattering of sunlight by air molecules. The blue light has
a frequency of about 7.5 x1014 Hz.
a. Calculate the wavelength, in nm, associated with this radiation
(a)
Strategy: We are given the frequency of an electromagnetic wave and asked to calculate the wavelength. Rearranging
Equation (7.1) of the text and replacing u with c (the speed of light) gives:
c

Solution: Substituting in the frequency and the speed of light (3.00  108 m/s) into the above equation, the wavelength is:
m
3.00  108
s  4.0  107 m  4.0  102 nm
 
14 1
7.5  10
s
 
Check: The wavelength of 400 nm calculated is in the blue region of the visible spectrum as expected.
b. Calculate the energy, in joules, of a single photon associated with this frequency
(b)
Strategy: We are given the frequency of an electromagnetic wave and asked to calculate its energy. Equation (7.2) of the
text relates the energy and frequency of an electromagnetic wave.
E  h
Solution: Substituting in the frequency and Planck's constant (6.63  1034 Js) into the above equation, the energy of a
single photon associated with this frequency is:
1

E  h  (6.63  1034 J  s)  7.5  1014   5.0  1019 J
s

Check: We expect the energy of a single photon to be a very small energy as calculated above, 5.0  1019 J.
A photon has a frequency of 6.0 x 104 Hz.
a. Convert this frequency into wavelength (nm). Does this frequency fall in the visible region?
3.
(a)
 
c
3.00  108 m/s

 5.0  103 m  5.0  1012 nm

6.0  104 /s
The radiation does not fall in the visible region; it is radio radiation. (See Figure 7.4 of the text.)
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b. Calculate the energy (in joules) of this photon
(b)
E  h  (6.63  1034 Js)(6.0  104 /s)  4.0  1029 J
c. Calculate the energy (in joules) of 1 mole of photons all with this frequency
(c)
Converting to J/mol: E 
4.0  1029 J 6.022  1023 photons

 2.4  105 J/mol
1 photon
1 mol
What is the wavelength, in nm, of radiation that has an energy content of 1.0 x 103 kJ/mol? In which
region of the electromagnetic spectrum is this radiation found?
4.
The energy given in this problem is for 1 mole of photons. To apply E  h, we must divide the energy by Avogadro’s
number. The energy of one photon is:
E 
1.0  103 kJ
1 mol
1000 J


 1.7  1018 J/photon
23
1 mol
1
kJ
6.022  10 photons
The wavelength of this photon can be found using the relationship, Ε 
hc
.

m

(6.63  1034 J  s)  3.00  108 
hc
1 nm
s 

 

 1.2  107 m 
 1.2  102 nm
18
9
E
1.7  10 J
1  10 m

The radiation is in the ultraviolet region (see Figure 7.4 of the text).
5.
When copper is bombarded with high-energy electrons, X rays are emitted. Calculate the energy (in
joules) associated with the photons if the wavelength of the X rays is 0.154 nm.
E  h 
hc
(6.63  1034 J  s)(3.00  108 m/s)

 1.29  1015 J

(0.154  109 m)
A particular form of electromagnetic radiation has a frequency of 8.11 x 1014 H.
a. What is its wavelength in nanometers? In meters?
6.
(a)
 
c

m
s  3.70  107 m  3.70  102 nm
 
14 1
8.11  10
s
3.00  108
b. To what region of the electromagnetic spectrum would you assign it?
(b)
Checking Figure 7.4 of the text, you should find that the visible region of the spectrum runs from 400 to 700 nm. 370
nm is in the ultraviolet region of the spectrum.
c. What is the energy (in joules) of one quantum of this radiation?
(c)
E  h. Substitute the frequency () into this equation to solve for the energy of one quantum associated with this
frequency.
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1

E  h  (6.63  1034 J  s)  8.11  1014   5.38  1019 J
s

Page 3 of 3