Practice Problems 13
Chapter 7
CHE 151
Graham/07
1.)
A laser emits light of frequency 4.74 x 1014 sec-1. What is the wavelength of
the light in nm?
 = c = 2.998 x 108 m x 1 s
x 1 nm = 6.32 x 102 nm

s
4.74 x 1014
10-9m
2.)
A certain electromagnetic wave has a wavelength of 625 nm.
a.) What is the frequency of the wave?
625 nm x 10-9 m = 6.25 x 10-7 m
1 nm
 = c = 2.998 x 10 m/s

6.25 x 10-7 m
8
 = 4.80 x 1014 s-1
b.) What region of the electromagnetic spectrum is it found?
Visible Region (~400 – 750 nm)
c.) What is the energy of the wave?
E = h= (6.626 x 10-34 J.s)(4.80 x 1014 s-1) = 3.18 x 10-19 J
3.)
How many minutes would it take a radio wave to travel from the planet
Venus to Earth? (Average distance from Venus to Earth = 28 million miles).
(Note: All electromagnetic travels at the speed of light in a vacuum)
2.8 x 107 mi
4.5 x 1010 m
4.)
x 1 km x
0.6214 mi
103 m = 4.5 x 1010 m
1 km
x 1s
x
8
2.998 x 10 m
1 min =
60 s
2.5 min
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 x 1014 Hz.
a.) Calculate the wavelength, in nm, associated with this radiation.
1 Hz = 1 s-1
 = c = 2.998 x 10 m x 1 nm = 4.0 x 10 nm
7.5 x 1014 s-1
10-9 m
8
2
b.) Calculate the energy, in joules, of a single photon associated with this
frequency.
E = h= (6.626 x 10-34 J.s)(7.5 x 1014 s-1) = 5.0 x 10-19 J
What is E in joules for an atom that releases a photon with a wavelength of
3.2 x 10-7 meters?
Eatom = Ephoton = hhc
5.)


E = (6.626 x 10-34 J.s)( 2.998 x 108 m/s) = 6.2 x 10-19 J
3.2 x 10-7 m
6.)
Calculate the frequency (Hz) and wavelength (nm) of the emitted photon
when an electron drops from the n=4 to n=2 state.
E = RH 1 - 1 = (2.179 x 10-18J) 1 - 1 = 2.179 x 10-18 - 2.179 x 10-18
ni2 nf2
42 22
16
4
E = h

 = E
= 1.362 x 10-19 - 5.448 x 10-19 = -4.086 x 10-19J = E
 = 4.086 x 10-19J = 6.167 x 1014Hz = 
6.626 x 10-34 J.s
h
= c =

7.)
2.998 x 108 m/s x 1 nm
6.167 x 1014 s-1
10-9m
=
486.1 nm
An electron in the hydrogen atom makes a transition from an energy state of
principal quantum numbers ni to the n = 2 state. If the photon emitted has a
wavelength of 434 nm, what is the value of ni?
Eatom = Ephoton = hhc = (6.626 x 10-34 J.s)( 2.998 x 108 m/s) = -4.58 x 10-19 J

434 x 10-9m
(negative number because
it is an emission process)
E = RH 1 - 1 = -4.58 x 10-19J = (2.179 x 10-18J)
ni2 nf2
-4.58 x 10-19J = 1 - 0.250 (keep 3 sf)
2.179 x 10-18J
ni2
ni =
8.)
1 - 1
ni2 22
-0.210 + 0.250 = 1 = 0.040
1
√ 0.040
ni2
ni = 5
Protons can be accelerated to speeds near that of light in particle
accelerators. Estimate the deBroglie wavelength (in nm) of such a proton
moving at 2.90 x 108 m/s. (mass of a proton = 1.673 x 10-27 kg).
1 J = 1 kg . m2
s2
=h
mu
h = 6.626 x 10-34 J.s x 1 kg . m2 = 6.626 x 10-34 kg . m2
s2
s
 =
6.626 x 10-34 kg . m2/s
= 1.37 x 10-15 m
-27
8
(1.673 x 10 kg)(2.90 x 10 m/s)
1.37 x 10-15 m x 1nm = 1.37 x 10-6 nm
10-9m
9.)
Calculate the deBroglie wavelength (in nm) of a 3000. lb automobile traveling
at 55 mi/hr.
55 mi x 1 km x 103 m x 1 hr x 1 min = 25 m
1 hr 0.6214 mi 1 km
60 min 60 sec
s
3000. lb x 1 kg = 1361 kg
2.2046 lb
 = h = 6.626 x 10-34 kg . m2/s = 1.9 x 10-38 m x 1 nm = 1.9 x 10-29 nm
mu
(1361 kg)(25 m/s)
10-9 m
10.)
What are the possible values of l for an electron with n=3?
l = (0….n-1)
11.)
l = 0,1,2
For the following subshells give the values of the quantum numbers (n, l and
ml) and the number of orbitals in each subshell.
(a) 4p
n=4
l=1
ml = -1,0,+1
(b) 3d
n=3
l =2
ml = -2,-1,0,+1,+2
(c) 3s
n=3
l=0
ml = 0
(3 p orbitals)
(5 d orbitals)
(1 s orbital)
(d) 5f
n=5
l=3
ml = -3,-2,-1,
0,+1,+2,+3
(7 f orbitals)
12.)
13.)
For each of the following, give the subshell designation, the allowable ml
values, and the number of orbitals.
(a) n = 2, l = 0
2s, ml = 0, (1 orbital)
(b) n = 3, l = 2
3d, ml = -2,-1,0,+1,+2 (5 orbitals)
(c) n = 5, l = 1
5p, ml = -1,0,+1
(3 orbitals)
Are the following quantum number combinations allowed? If not, show two
ways to correct them.
(a) n = 1; l = 0; ml = 0 yes. 1s
(b) n = 2; l = 2; ml = +1
No
n = 3; l = 2; ml = +1 or n = 2; l = 1; ml = +1
(c) n = 7; l = 1; ml = +2
n = 7; l = 1; ml = +1
(d) n = 3; l = 1; ml = -2
n = 3; l = 1; ml = -1
No
or n = 7; l = 2; ml = +2
No
or
n = 3; l = 2; ml = -2
14.)
The energy required to remove an electron from metal X is
E = 3.31 x 10-20J. Calculate the maximum wavelength of light that can
photo eject an electron from metal X.
E = hc 

= hc = (6.626 x 10-34 J.s)( 2.998 x 108 m/s)
E
3.31 x 10-20 J
= 6.00 x 10-6 m x 1nm =
10-9m
15.)
If an electron has a velocity of 5.0 x 105 m/s, what is its wavelength in m?
1 J = 1 kg . m2
s2
=h
mu
m = mass of electron = 9.109 x 10-28g x 1 kg = 9.109 x 10-31 kg
103
-34 .
.
2
h = 6.626 x 10 J s x 1 kg m = 6.626 x 10-34 kg . m2
s2
s
 =
16.)
6.00 x 103 nm
6.626 x 10-34 kg . m2/s
(9.109 x 10-31 kg)(5.0 x 105 m/s)
= 1.5 x 10-9 m
The laser used to read information from a compact disk has a wavelength of
780 nm. What is the energy associated with one photon of this radiation?
Ephoton = hc = (6.626 x 10-34 J.s)( 2.998 x 108 m/s) = 2.55 x 10-19 J
780 x 10-9 nm
17.)
The retina of a human eye can detect light when radiant energy incident on it
is at least 4.0 x 10-17 J. For light of 600 nm wavelength, how many photons
does this correspond to?
1.) Determine the energy of 1 photon:
Ephoton = hc = (6.626 x 10-34 J.s)( 2.998 x 108 m/s) = 3.31 x 10-19 J/ photon
600 x 10-9 nm
2.) Calculate # photons needed to produce given amount of energy:
4.0 x 10-17 J x 1 photon = 1.2 x 102 photons
3.31 x 10-19 J