Physics 112
Homework 14 (Ch27 &28)
Due July 31
1. 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.
5.0  10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
v  1.5  106 m s.
(c) With v  0.005 c, we can calculate kinetic energy classically:
K


 12 mv 2  12 9.11  1031 kg 1.5  106 m s

2
 9.7  1019 J,
9.7 10

19
J
 6.0eV
1.6  10 J / eV
This is the energy gained by an electron as it is accelerated through a potential difference of
which converted to electron-volts equals

19

6.0V.
2. 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
1
 2m0 K 2
29.1110
6.63 1034 J  s
31


kg 2450eV  1.6 10 C / e
Method 2

hc
 2m0 c 2 K 
1
2
;

19

1/ 2
 2.48 1011 m  2.48 102 nm
1243eV  nm
20.51110 eV 2450eV 
This is the maximum possible resolution.
6
1/ 2
 2.48 102 nm
Physics 112
Homework 14 (Ch27 &28)
Due July 31
3. 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

 2m0 K 
 6.63  10 J s 
kg   0.025eV  1.6  10
34
h
1
2

 2 1.67  1027

h
hc
1243eV  nm

1 
2
p 2m0cK 
2 939.57 106 eV 0.025eV



1
2
19
J eV  
1
2
 1.81  1010 m.
 0.181nm  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
y 

nL
.
d

L 1.0m  1.81  10 10 m

 3.6  10 7 m
3
d
0.50  10 m
4. 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
 
5. 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
Physics 112
x 
Homework 14 (Ch27 &28)
Due July 31
1.055  1034 J s 
ÿ

 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.
6. For n  6, what values can l have?
Solution
The value of
can range from 0 to n  1. Thus for n = 6, we have
 0, 1, 2, 3, 4, 5.
7. 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.
8. 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 .
9. 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 .
Physics 112
Homework 14 (Ch27 &28)
Due July 31
(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.
10. 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
L  ÿ 
(d) For each

62
 0.378eV.
 0. The magnitude of the angular momentum depends on
 1  1.055  1034 J s   0  0  1  0.
1
2
1
2
the value of m can range from  to  : m  0.
only: