CC.

5111 Extra Problems for Exam I solutions

Question 1
Photons of wavelength 315 nm or less are needed to eject electrons from a surface of
electrically neutral cadmium. What is the energy barrier that electrons must overcome?
The energy barrier is the work function:

)(

6.626 x 10-34 J s 2.9979 x 10 8 m s-1

hc
=
=
= 6.306 x 10-19 J = 6.31 x 10-19 J
-9

315 x 10 m
Question 2
Calculate the energy (in J) of a photon whose period is 5 x 10-15 s.
We can calculate the frequency of the photon by the following equation:
1
period = = 5 x 10-15 s

= 2 x 1014 s-1
Substitute the frequency into the equation for energy:

)(

E = h = 6.626 x 10 34 J s 2 x 1014 s-1 = 1.325 x 10 19 J = 1 x 10 19 J (correct sig figs)

Question 3
(a) Calculate the momentum of a photon whose wavelength is that of a He-Ne
laser, 6.330 x 10-7 m.
Calculate the photon momentum from its wavelength via the de Broglie relationship:
h 6.626076 x 10 34 J s
p= =
= 1.04677 x 10-27 kg m s-1 = 1.047 x 10-27 kg m s-1
-7

6.330 x 10 m
(b) Calculate the momentum of an electron at 100. eV.
First convert the energy of the electron from eV to J:

100. eV 1.60218 x 10-19 J/eV = 1.60218 x 10-17 J

We can calculate its momentum from its energy;

p = 2mE = (2) 1.6022 x 10-17 J

) ( 9.109 x 10-31 kg ) = 5.403 x 10-24 kg m s-1 = 5.40 x 10-24 kg m s-1

Question 4
A free electron moves with a velocity v = 1.00x106 m/s before being captured by a
hydrogen nucleus. Eventually, it lands in the 1s orbital. What is the change in its
energy?
First let us determine the amount of energy contained in the free electron.

2
1
! 1$
m e v2 = # & 9.1094 x 10 31kg 1.00 x 10 6 m s-1 =4.5547 x 10-19 J
" 2%
2
We can then determine the energy of the 1s orbital

KE =

)(

12
E1 = 2 (2.18 x 10 18 J) = 2.18 x 10 18 J
1
The differences in the energy levels is just the difference between the two energy values:
E=E Final E inital = -2.18 x 10-18 J 0.45547 x 10-18 J = -2.6354 x 10-18 J = -2.64 x 10-18 J (correct sig figs)

The energy difference between the two states is -2.64 x 10-18 J. The energy is observed as
emission of a photon as the electron is attracted into the 1s energy level. This problem
is the reverse of the photoelectric effect:
4.555 x 10-19 J
2.64 x 10-18J
2.18 x 10-18 J

Question 5
Compute the difference in energy between the fourth and fifth shells of an H atom.
Is the comparable spacing more or less in a He+ ion?
We can calculate the difference by using the equation for the hydrogen H;
)" 1 % " 1 % ,
E 5-4 = + $ 2 ' $ 2 ' . 2.18 x 10-18 J = 4.905 x 10-20 J = 4.91 x 10-20 J (correct sig figs)
*# 4 & # 5 & -

Since He+ ion is also a one-electron system, the energy difference can be calculated by
using the equation for the H atom, but notice that we have to take into consideration
that for He+, the atomic number is 2, which is double that of H. We can substitute into
the equation:
) " 22 % " 22 % ,
E 5-4 = + $ 2 ' $ 2 ' . 2.18 x 10-18 J = ( 0.25 0.16 ) 2.18 x 10-18 J = 1.96 x 10-19 J (correct sig figs)
+* # 4 & # 5 & .-

The energy difference in the He+ ion is greater than that of an H atom.
Question 6
The ionization energy of a certain one-electron atom in its ground state is 3.28 x 104
kJ/mol. How many protons are contained in the nucleus?
Convert the IE from a value for an entire mole of the unknown, to a single atom,
! 3 $
1 mol
$
4
-1 10 J !
IE= 3.28 x 10 kJ mol #
=5.4467 x 10-17 J
& #"
&
23
" kJ % 6.0221 x 10 atoms %

We also know that IE = -En and that an equation for En is,

Z2
E n = (2.18 x 1018 J) 2
n
So rearranging the equation, we can come up with the following relationship,
Z2
-5.4467 x 10 -17 J = (2.18 x 1018 J) 2

n
2
Z
24.984 = 2
n
Z
= 4.9984 5
n
We know that n must equal 1, since we are told that it is a one electron atom in the ground
state. Therefore, the atomic number of this atom is 5 (Boron). There are 5 protons in the

nucleus.
Question 7
At r = 0, the PROBABILITY DENSITY of an H atom wavefunction has a non-zero
value. In addition, the probability density associated with this wavefunction is zero
for at least one non-zero value of r. Of the possible wavefunctions from n = 1
through n = 4 and from = 0 through = 1, name all the wavefunctions that are
consistent with these probability densities.
The only wavefunctions with non-zero probability density at r = 0 are the s wavefunctions.
Since there is at least one node, we know it cannot be the 1s wavefunction. We are limited to
wavefunctions with n 4, the wavefunctions fitting this description are the 2s, 3s, and 4s
wavefunctions.
Question 8
An atom of sodium has the electron configuration [Ne] 6s1. Explain how this is
possible.
This atom of sodium is in an excited state. It can lose energy in a variety of ways to end
up ultimately in its ground state, which is represented by[Ne] 3s1.
Question 9
The most probable value of r for the 3p state is smaller than that for the 3s state. Why
is the energy of the 3s state lower than that of the 3p state in a multielectron atom?
If you look at the RPD plots of the 3s and the 3p orbitals, they look something like this:

3p

RPD

3s

If you look at the probability of finding the electron close to the nucleus in the 3s orbital,
there is a small probability that it is close to the nucleus, whereas there is almost no
probability that a 3p is close to the nucleus. This small probability produces, on
3

average, a higher Zeff for 3s than for 3p. Since the binding energy is proportional to the
square of Zeff, the binding energy is larger for the 3s electron than the 3p electron.
Question 10
The photoelectron spectrum of Mg is measured using X-ray photons of energy 600.0 x
10-18 J. Electrons are emitted from the Mg atoms with the following kinetic energies
(in units of 10-18 J): 598.8, 591.4, 585.3, 391.0.
(a) Write the ionization processes responsible for each of these kinetic energies.
The electronic configuration of Mg is 1s22s22p63s2. The orbitals in order of
increasing ionization energies are 3s2 < 2p6 < 2s2 < 1s2. The higher the ionization
energy, the lower the kinetic energy, so we can assign the kinetic energies to the
various ionization energies.
Mg(1s22s22p6-3s2) + photon Mg+ (1s12s22p63s2 ) + e- (KE = 391.0 x 10-18J)
Mg(1s22s22p6-3s2) + photon Mg+ (1s22s12p63s2 ) + e- (KE = 585.3 x 10-18J)
Mg(1s22s22p6-3s2) + photon Mg+ (1s22s22p53s2 ) + e- (KE = 591.4 x 10-18J)
Mg(1s22s22p6-3s2) + photon Mg+ (1s22s22p63s1 ) + e- (KE = 598.8 x 10-18J)
(b) Calculate the binding energies for the filled orbitals of Mg
We can determine the binding energies of the various electrons by remembering
the following relationship: E i = KE + Ionization Energy (or )
For the 1s electron
600.0 x 1018 J = 391.0 x 10 -18 J + IE

(c)

IE = 209.0 x 10 -18 J
The binding energy for the 1s electron is -2.090 x 10-16 J.
For the 2s electron
600.0 x 1018 J = 585.3 x 10 -18 J + IE
IE = 14.7 x 10 -18 J
The binding energy for the 2s electron is -1.47 x 10-17 J
For the 2p electron
600.0 x 1018 J = 591.4 x 10 -18 J + IE
IE = 8.6 x 10 -18 J
The binding energy for the 2p electron is -8.6 x 10-18 J.
For the 3s electron
600.0 x 1018 J = 598.8 x 10 -18 J + IE
IE = 1.2 x 10 -18 J
The binding energy for the 3s electron is -1.2 x 10-18 J.

Al+ is isoelectronic with Mg. Do you expect the binding energies in Al+ to be
weaker or stronger than in+ Mg? Explain.
The orbital energies of Al should be lower than that of Mg because Al+ has a
higher Zeff than for Mg. This means that the electrons are going to be bound more
tightly to the nucleus, hence the stronger binding energies.

Question 11:
Consider two photoelectric effect experiments on Cs metal. One experiment uses a
1.00 102 W green lamp with a wavelength of 530.6 nm and the other uses a 5.00
102 W blue lamp with a wavelength of 460.5 nm. (1W = 1 J/s)

a) Calculate the work function of Cs if the 1.00 102 W green lamp with a
wavelength of 530.6 nm produces electrons with an energy of 7.01 10-20 J.
First let us draw a picture with all the information that we have:
light with
wavelength of
530.6 nm

electrons are ejected

with Kinetic energy of
7.01 10-20 J

Cs metal

We know that the total energy of the lamp is 1.00 102 W, but it is delivered in
packets of energy with a wavelength of 530.6 nm (530.6 x 10-9m).
We can calculate the work function of the Cs metal using the equation for
photoelectric effect:
E i = + KE
hc
= KE

6.62608 x 10 -34 Js)(2.9979 x 10 8 m s1 )

(
=
7.01 x 10 -20 J
530.6 x 10 -9 m
= 3.7437 x 10 -19 J - 0.701 x 10 -19 J
= 3.0427 x 10 -19 J
= 3.043 x 10 -19
The work function of Cs metal is 3.043 x 10-19J.

b) Calculate
how many photons per s are being emitted by the 5.00 102 W blue
lamp with a wavelength of 460.5 nm.
The energy per photon can be calculated from the wavelength and the total energy:
-34
8
1
hc (6.62608 x 10 Js)(2.9979 x 10 m s )
E photon =
=

460.5 x 10 -9 m
= 4.3136 x 1019 J per photon

We can now calculate the number of photons per second:

E
5.00 x 10 2 J s -1
# of photons = total =
E photon 4.3136 x 1019 J (photon) -1

= 1.159 x10 21 photons s-1

= 1.16 x10 21 photons s-1
From the blue lamp, there are 1.16 x 1021 photons/sec emitted.

c) If the Cs metal is exposed to the 1.00 102 W green lamp with a wavelength of
530.6 nm for 1.0 min, calculate how long must the Cs metal be exposed to the
5.00 102 W blue lamp with a wavelength of 460.5 nm to eject the same
number of electrons.
First, we need to calculate the number of electrons ejected by the green lamp. In (a)
we calculated the energy of each photon, Ei. Since the incident energy, Ei, is greater
than the work function, electrons will be emitted when Cs is exposed to the green
light. In fact the number of electrons emitted is equal to the number of photons
coming from the lamp (answer to part c):
E
1.00 x 10 2 J s -1
# of photons = total =
E photon 3.7437 x 1019 J (photon) -1
= 2.671 x10 20 photons s-1
= 2.671 x10 20 electrons s-1
We can now calculate the total number of electrons emitted in 60 seconds to be:
" 2.671 x10 20 electrons %
total # of electrons ejected = $
'(60 s)
s

#
&
= 1.60 x 10 22 electrons ejected in 1 minute
with the green lamp

Using our answer from part (c):

1.16 x10 21 photons s-1 = 1.16 x10 21 electrons s-1
we know that we want to eject the same number of electrons as with the green
And
lamp (1.60 x 1022 electrons in 1 minute), so we can calculate the time it will take to

ejected the same

number of electrons:
# of electrons
time to eject 1.60 x 1022 electrons =
electrons ejected per second
1.60 x 10 22 electrons
1.16 x10 21 electrons s-1
= 13.8 s
= 14 s
It will take 14 seconds to eject the same number of electrons using the blue lamp
=

Question 12:
Draw the radial probability distribution for an electron in a 4p orbital. The relative
magnitudes of the extrema and nodes must be correct, if there are multiple extrema
and nodes. Label your axes. Indicate the origin on both the y and x axes.

# of radial nodes = n -1- l

= 4 11
=2

RPD graph:

Question 13:
a) Write the ground state electron configuration of Cd+. You can use the inert gas
configuration as a means to abbreviate the configuration.
The ground state electron configuration of Cd is [Kr] 5s24d10. We can rewrite this to
reflect that filled 4d orbitals are lower in energy than 5s orbitals: [Kr]4d105s2. If we are
going to remove an electron it must be removed from the s orbital, so the electron
configuration of Cd+ is [Kr]4d105s1.
b) Consider the ni = 2 to nf = 3 transition in C+5 and in O+7. Does this transition
in O+7 require a photon with the same momentum, a smaller momentum or a
Z 2 R H "$ 1
1 %'
larger momentum as the transition in C+5? (Refer to =
).

h $# n 2i n2f '&
We need to first think about how momentum and frequency are related. Looking at
h h
the deBroglie equation: p = =
, we can see that momentum and frequency are
c
directly proportional.
Now we can determine which has a higher frequency:
2
'
6) (2.18 x 10 -18 J) $ 1
(
1 )
+5

&
For C : C+5 =

6.62608 x 10 -34 Js & (2) 2 ( 3) 2 )

%
(
= 1.65 x 1016 Hz
+7

For O : O +7 =

(8)

(2.18 x 10

-18

6.62608 x 10 -34

'
J) $ 1
1 )
&

Js & (2) 2 ( 3) 2 )
%
(

= 1.52 x 1018 Hz

Since the frequency of the O+7 is larger, then the transition in O+7 will require more energy
and therefore the O+7 transition will require a photon with greater momentum than in C+5.
Question 14: Which of the following has the largest atomic radius: S2-, Cl, Cl-, K+, or
S? Explain very briefly.
S2-. It has a net -2 charge that cannot be held close to the nucleus with its charge deficit.