Physics 43 Chapter 44: HW 4
Serway 7th Edition Problems: 5,7,9,10,19,20,21,28,34,37,42,44
5.
(a) Use energy methods to calculate the distance of closest approach for a head-on collision
between an alpha particle having an initial energy of 0.500 MeV and a gold nucleus (197Au) at rest. (Assume
the gold nucleus remains at rest during the collision.) (b) What minimum initial speed must the alpha
particle have to get as close as 300 fm?
P44.5
(a)
The initial kinetic energy of the alpha particle must equal the electrostatic potential energy
at the distance of closest approach.
Ki  U f 
keqQ
rmin
9
2
2
19
k qQ  8.99  10 N  m C   2 79 1.60  10 C 
 e

 4.55  1013 m
Ki
 0.500 M eV  1.60  1013 J M eV 
2
rmin
(b)
Since Ki 
k qQ
1
m vi2  e
,
2
rmin
2  8.99  109 N  m 2 C2   2 79 1.60  1019 C 
2keqQ
vi 

 6.04  106 m s
m rmin
 4.00 u  1.66  1027 kg u  3.00  1013 m 
2
P44.7 A star ending its life with a mass of two times the mass of the Sun is expected to
collapse, combining its protons and electrons to form a neutron star. Such a star could be
thought of as a gigantic atomic nucleus. If a star of mass 2 × 1.99 × 10 30 kg collapsed into
neutrons (mn = 1.67 × 10–27 kg), what would its radius be? (Assume that r = r0A1/3.)
Solution
The number of nucleons in a star of two solar masses is
A
Therefore


2 1.99  1030 kg
27
1.67  10

kg nucleon
r  r0A 1 3  1.20  1015 m
 2.38  1057 nucleons.
 2.38 10 
57 1 3
 16.0 km
. !!!!!
9. Calculate the binding energy per nucleon for (a) 2H, (b) 4He, (c) 56Fe, and (d) 238U.
Eb = (Zmp + Nmn – MA) x 931.494 MeV/u
Using atomic masses as given in the table in the text,
2.014 102  11.008 665  11.007 825
(a)
For 21 H :
(b)
For 42 He :
(c)
For 56
26 Fe :
2
Eb
 931.5 MeV 
  0.001194 u  
  1.11 MeV nucleon
A
u


2 1.008 665  2 1.007 825  4.002 603
4
Eb
2
 0.007 59 u c  7.07 MeV nucleon
A
301.008 665  261.007 825  55.934 942  0.528 u
Eb 0.528

 0.009 44 u c2  8.79 MeV nucleon
A
56
(d)
1461.008 665  921.007 825  238.050 783  1.934 2 u
For 238
92 U :
Eb 1.934 2

 0.008 13 u c2  7.57 MeV nucleon
A
238
10.
The iron isotope 56Fe is near the peak of the stability curve. This is why iron is generally prominent
in the spectrum of the Sun and stars. Show that 56Fe has a higher binding energy per nucleon than its
neighbors 55Mn and 59Co. Compare your results with Figure 44.5.
M  ZmH  Nmn  M
*P44.10
Nuclei
Z
N
M in u
M in u
Eb
in MeV
A
8.765
8.790
8.768
Eb M  931.5

A
A
25
30
54.938 050
0.517 5
26
30
55.934 942
0.528 46
27
32
58.933 200
0.555 35
E
 56 Fe has a greater b than its neighbors. This tells us finer detail than is shown in Figure 44.5.
A
55
Mn
Fe
59
Co
56
A sample of radioactive material contains 1.00 × 1015 atoms and has an activity of
6.00 × 1011 Bq. What is its half-life?
19
dN
  N
dt

T1 2 
ln 2


1  dN 
15
6.00  1011  6.00  104 s1

  1.00  10
N  dt 

so
 1.16  103 s   19.3 m in

20. The half-life of 131I is 8.04 days. On a certain day, the activity of an iodine-131 sample
is 6.40 mCi. What is its activity 40.2 days later?
5
 1
  6.40 mCi   e ln 2    6.40 mCi   5   0.200 mCi
2 
P44.20 R  R0 e  t   6.40 mCi  e 
 ln 2 8.04 d  40.2 d 
A freshly prepared sample of a certain radioactive isotope has an activity of 10.0
mCi. After 4.00 h, its activity is 8.00 mCi. (a) Find the decay constant and half-life. (b)
How many atoms of the isotope were contained in the freshly prepared sample? (c) What
is the sample’s activity 30.0 h after it is prepared?
21
Solution
(a)
R  R0e t ,
From
1
R 


1
 10.0
2 1
5 1
  ln  0   
 ln 
  5.58  10 h  1.55  10 s
t  R   4.00 h   8.00
T1 2 
R0

 12.4 h
10.0  103 Ci 3.70  1010

1 Ci
1.55  105 s 
s
13
  2.39  10 atom s

(b)
N0 
(c)
R  R0e t  10.0 m Ci exp 5.58  102  30.0  1.88 m Ci


ln 2

28. Identify the missing nuclide or particle (X):
(a) X 
(b)
215
84
109
48
(e)
14
7
Ni  
X=
Po  X  
(c) X 
(d)
65
28
55
26
65
28
X
Ni
211
82
Pb

Fe  e  v
X=
Cd  X  109
47 Ag  v
N  42 He  X 
17
8
O
55
27
Co
X=
X=
0
1
1
1
e
H

28 solution:
(a)
A gamma ray has zero charge and it contains no protons or neutrons. So for a gamma ray Z  0
and A  0 . Keeping the total values of Z and A for the system conserved then requires Z  28 and A  65

for X. With this atomic number it must be nickel, and the nucleus must be in an exited state, so it is 65
28 Ni .
  42 He has
(b)
so for X
for Pb
Z  2 and
A4
we require
Z  84  2  82
and
A  215  4  211 , X  211
82 Pb
(c)
A positron e  01 e has charge the same as a nucleus with Z  1 . A neutrino 00 has no charge.
Neither contains any protons or neutrons. So X must have by conservation Z  26  1  27 . It is Co.
55
And A  55  0  55 . It is 27
Co .
Similar reasoning about balancing the sums of Z and A across the reaction reveals:
(d)
0
1
e
1
(e)
1 H (or p). Note that this process is a nuclear reaction, rather than radioactive decay. We can solve it
from the same principles, which are fundamentally conservation of charge and conservation of baryon
number.
34.
Determine which decays can occur spontaneously:
Ca  e  
(a)
40
20
(b)
98
44
144
60
(c)
40
19
K
Ru  42 He  94
42 Mo
4
Nd  2 He  140
58 Ce
P44.34
(a)
For e decay,
Q   M X  M Y  2me  c2  39.962 591 u  39.963 999 u  2  0.000 549 u    931.5 MeV u 
Q  2.33 MeV
Since Q  0 , the decay cannot occur
(b)
spontaneously.
For alpha decay,
Q   M X  M   M Y  c2   97.905 287 u  4.002 603 u  93.905 088 u   931.5 MeV u 
Q  2.24 MeV
Since Q  0 , the decay cannot occur
(c)
spontaneously.
For alpha decay,
Q   M X  M   M Y  c2  143.910 083 u  4.002 603 u  139.905 434 u   931.5 MeV u 
Q  1.91 MeV
Since Q  0 , the decay can occur
spontaneously.
37. Indoor air pollution. Uranium is naturally present in rock and soil. At one step in its series of radioactive
decays, 238U produces the chemically inert gas radon-222, with a half-life of 3.82 days. The radon seeps out
of the ground to mix into the atmosphere, typically making open air radioactive with activity 0.3 pCi/L. In
homes 222Rn can be a serious pollutant, accumulating to reach much higher activities in enclosed spaces. If
the radon radioactivity exceeds 4 pCi/L, the Environmental Protection Agency suggests taking action to
reduce it, such as by reducing infiltration of air from the ground. (a) Convert the activity 4 pCi/L to units of
becquerel per cubic meter. (b) How many 222Rn atoms are in one cubic meter of air displaying this activity?
(c) What fraction of the mass of the air does the radon constitute?
P44.37
N
(b)
 4.00  1012 Ci   3.70  1010 Bq   1.00  103 L 
3
4.00 pCi L  


  148 Bq m
1L
1 Ci
1 m3




(a)
 T1 2 
3  3.82 d   86 400 s 
7
3
 R
   148 Bq m  

  7.05 10 atoms m

ln
2
ln2
1
d





R
1 mol

  222 g 
14
3
mass   7.05 107 atoms m 3  

  2.60  10 g m
23
 6.02  10 atoms  1 mol 
Since air has a density of 1.20 kg m 3 , the fraction consisting of radon is
(c)
fraction 
42.
2.60  1014 g m 3
 2.17  1017
1 200 g m 3
A beam of 6.61-MeV protons is incident on a target of
p
27
( 14
27
13
Al 
27
14
27
13
Al . Those that collide produce the reaction
Si  n
Si has mass 26.986 705 u.) Ignoring any recoil of the product nucleus, determine the kinetic energy of
the emerging neutrons.
P44.42 Neglect recoil of product nucleus (i.e., do not require momentum conservation for the system of colliding
particles). The energy balance gives Kemerging  Kincident  Q . To find Q:
Q   M H  M Al    M Si  mn   c2
Q   1.007 825  26.981 539   26.986 705  1.008 665  u  931.5 MeV u   5.59 MeV
Thus,
44.
(a) Suppose
10
5
Kemerging  6.61 MeV  5.59 MeV  1.02 MeV .
B is struck by an alpha particle, releasing a proton and a product nucleus in the
reaction. What is the product nucleus? (b) An alpha particle and a product nucleus are produced when
is struck by a proton. What is the product nucleus?
P44.44
(a)
10
5
B  42 He  136 C  11 H
The product nucleus is
(b)
13
6
13
6
C .
10
5
B .
C  11 H  105 B  42 He
The product nucleus is
13
6
C