Homework 15
Ch27: P 41, 47, 49, 55; Ch28: P 1, 5, 9, 13, 15; 21, 23, 25;
Problems (Ch27):
41. (II) An electron has a de Broglie wavelength   5.0  1010 m. (a) What is its
momentum? (b) What is its speed? (c) What voltage was needed to accelerate it to
this speed?
Solution
(a) We find the momentum from
p
h


6.63  1034 J s
 1.3  1024 kg m s.
10
5.0  10 m
(b) We find the speed from
p
h
1.3  10 24 kg  m / s
v 
,
v
 1.5  10 6 m / s ,
31
m m
9.11  10 kg
(c) With v  0.005 c, we can calculate
KE

KE
classically:


 12 mv2  12 9.11  1031 kg 1.5  106 m s
2
v  1.5  106 m s.
 9.7  1019 J,
 9.7  10 
19
which converted to electron-volts equals
1.06  10
19
J eV 
 6.0 eV.
This is the energy gained by an electron as it is accelerated through a potential difference
of
6.0V.
*47. (II) Electrons are accelerated by 2450 V in an electron microscope. What is the
maximum possible resolution?
Solution
The wavelength of the electron is


h

p
h
 2m0  KE  
1
2

hc
1
 2m0 c 2  KE  2
1243eV  nm
20.511 10 eV 2450eV 
6
1/ 2
 2.48  10 2 nm
This is the maximum possible resolution, as discussed in Section 27-9.
49. (I) How much energy is needed to ionize a hydrogen atom in the n  2 state?
Solution
To ionize the atom means removing the electron, or raising it to zero energy:
Eion  0  En 
13.6 eV   13.6 eV  
n2
22
3.4 eV.
55. (II) What wavelength photon would be required to ionize a hydrogen atom in the
ground state and give the ejected electron a kinetic energy of 10.0 eV?
Solution
The energy of the photon is
hf  Eion 
KE
 13.6 eV  10.0 eV  23.6eV.
We find the wavelength from

hc 1243eV  nm

 52.7nm
hf
23.6eV
Problems (Ch28):
1. (II) The neutrons in a parallel beam, each having kinetic energy 401 eV, are directed
through two slits 0.50 mm apart. How far apart will the interference peaks be on a
screen 1.0 m away? [Hint: First find the wavelength of the neutron.]
Solution
We find the wavelength of the neutron from

h

p
h
1
 2m0  KE   2
 6.63  10 J s 
kg   0.025eV  1.6  10
34

 2 1.67  1027

19
J eV  
1
2
 1.81  1010 m.
The peaks of the interference pattern are given by
d sin   n , n  1, 2, ... .
and the positions on the screen are
y  L tan .
For small angles, sin  tan , so we have
y
Thus the separation is
L 1.0m  1.81  10 10 m
y 

 3.6  10 7 m
3
d
0.50  10 m


nL
.
d
5. (I) An electron remains in an excited state of an atom for typically 10 8 s. What is the
minimum uncertainty in the energy of the state (in eV)?
Solution
We find the minimum uncertainty in the energy of the state from
34
ÿ 1.055  10 J s 
E 

 1.1  1026 J  6.6  108 eV.
8
t
10
s
 
9.
(II) An electron and a 140-g baseball are each traveling 150 m s measured to an
accuracy of 0.055%. Calculate and compare the uncertainty in position of each.
Solution
The uncertainty in the velocity is
 0.055 
v  
 150m s   0.0825m s.
 100 
For the electron, we have
1.055  1034 J s 

ÿ
x 

 1.4  103 m.
m v  9.11  1031 kg   0.0825m s 
For the baseball, we have
1.055  1034 J s   9.1  1033 m.
ÿ
x 

m v  0.140kg  0.0825m s 
The uncertainty for the electron is greater by a factor of 1.5  1029.
13. (I) For n  6, what values can l have?
Solution
The value of
15.
can range from 0 to n  1. Thus for n = 6, we have
 0, 1, 2, 3, 4, 5.
(I) How many electrons can be in the n  6, l  3 subshell?
Solution
The number of electrons in the subshell is determined by the value of . For each the
value of m can range from  to  , or 2  1 values. For each of these there are two
values of ms . Thus the total number for  3 is
N  2  2  1  2 2  3  1  14 electrons.
21. (II) If a hydrogen atom has ml  3, what are the possible values of n, l, and m s ?
Solution
The value of m can range from  to   so we have  3.
The value of can range from 0 to n  1. Thus we have n   1(minimum4).
There are two values of ms : ms   12 ,  12 .
23. (II) What is the full electron configuration in the ground state for elements with Z
equal to (a) 27, (b) 36, (c) 38? [Hint: See the periodic table inside the back cover.]
Solution
(a) We start with hydrogen and fill the levels as indicated in the periodic table:
1s 2 2s 2 2 p 6 3s 2 3 p 6 3d 7 4s 2 .
Note that the 4s 2 level is filled before the 3d level is started.
(b) For Z  36 we have
1s 2 2s 2 2 p 6 3s 2 3 p 6 3d 10 4s 2 4 p 6 .
(c) For Z  38 we have
1s 2 2s 2 2 p 6 3s 2 3 p 6 3d 10 4s 2 4 p 6 5s 2 .
Note that the 5s 2 level is filled before the 4d level is started.
25.
(II) A hydrogen atom is in the 6s state. Determine (a) the principal quantum
number, (b) the energy of the state, (c) the orbital angular momentum and its quantum
number l, and (d) the possible values for the magnetic quantum number.
Solution
(a) The principal quantum number is n  6.
(b) The energy of the state is
E6  
13.6eV    13.6eV  
n2
(c) The “s” subshell has
only:
L  ÿ 
(d) For each

62
 0.378eV.
 0. The magnitude of the angular momentum depends on
 1 2  1.055  1034 J s   0  0  1 2  0.
1
1
the value of m can range from  to  : m  0.