Homework 16
Ch30: P 1, 9, 29, 39, 45; Ch31: P 21, 31; Ch32: P 1, 9, 11
Problems (Ch30):
1.
(I) A pi meson has a mass of 139 Me V c 2 . What is this in atomic mass units?
Solution
139 MeV c 
2
m
We convert the units:
9.
(I) Estimate the total binding energy for
 931.5MeV uc 
40
20 Ca,
2
 0.149 u.
using Fig. 30–1.
Solution
From Figure 30–1, we see that the average binding energy per nucleon at
A  40 is 8.6MeV.
Thus the total binding energy for
40
 408.6MeV   340MeV.
Ca is
29. (II) How much energy is released in electron capture by beryllium:
7
4 Be

0
7
1 e  3 Li
 v?
Solution
For the electron capture 74 Be  01 e  73 Li  v, we see that if we add three electron masses
to both sides to use the atomic mass for Li, we use the captured electron for the atomic
mass of Be. We find the Q value from
Q  m  7 Be   m  7 Li  c2
  7.016928u    7.016003u  c 2  931.5MeV uc 2   0.861MeV.
39. (I) What fraction of a sample of
after 3.0 yr?
68
32 Ge,
whose half-life is about 9 months, will remain
Solution
We find the fraction remaining from
N  N0et ;
  0.693 3.0 yr 12mo yr  

9mo


N
 e  t  e 
N0
 0.0625.
45. (II) Calculate the mass of a sample of pure
40
2.0 105 s 1 . The half-life of 19
K is 1.28109 yr.
Solution
N

 2.0  10 5 s 1
t
T 1  1.28  10 9 yr
40
19 K
N
 N
t
ln 2

T1
2
  40 g / mol
with an initial decay rate of
 N  T12
N  

 t  ln 2
2
m?
We find the number of nuclei
9
7
5 1 1.28  10 yr 3.16  10 s / yr
N  2.0  10 s
 1.7  10 22 nuclei
ln 2


The mass is
m

1.17  10
22
nuclei   40.0g / mol 
6.02  1023 nuclei / mol

 0.77 g.
Problems (Ch31):
21. (II) How much mass of 238
92 U is required to produce the same amount of energy as
burning 1.0 kg of coal (about 3  10 7 J )?
Solution
We assume that an average of 200 MeV is released per fission of a uranium nucleus (see
Eq31-5).
0.238 kg
 1MeV   1nucleus  

7
238
3 107 J   1.601eV


  3.7 10 kg 92 U
19 
6
23
10 J  110 eV   200 MeV   6.02 10 nuclei 
31. (II) If a typical house requires 950 W of electric power on average, what minimum
amount of deuterium fuel would have to be used in a year to supply these electrical
needs? Assume the reaction of Eq. 31–8b.
Solution
The reaction of Eq. 31-8b consumes 2 deuterons and releases 3.27 MeV of energy. The
atomic
mass of Deuterium is 2.01g. The amount of energy needed is the power times the elapsed
time, and the energy can be related to the mass of deuterium by the reaction.
3
J
1eV


 1MeV   2 d   2.01  10 kg 
7 s 
950
1
yr
3.156

10










s
yr   1.60  1019 J  106 eV   3.27 MeV  6.02  1023 d 


 3.8  104 kg  0.38g
Problems (Ch32):
1.
(I) What is the total energy of a proton whose kinetic energy is 6.35 GeV?
Solution
The total energy is given by Eq. 26-7a.
E  m0 c  KE  0.94 GeV  6.35GeV  7.29 GeV
2
9.
(II) What is the wavelength (  maximum resolvable distance) of 7.0-TeV protons?
Solution
Because the energy of the protons is much greater than their rest mass, we have
KE  E  pc .
Combine that with Eq. 27-8 for the de Broglie wavelength.
E  pc ; p 
h

 E
hc

 
hc
E
 6.63 10 J s  3.0 10 m s   1.8 10

 7.0 10 eV 1.60 10 J eV 
34
8
19
12
19
m
11. (II) The 4.25-km-radius tunnel that will be used to house the magnets for the Large
Hadron Collider (LHC) calls for proton beams of energy 7.0 TeV. What magnetic
field will be required?
Solution
Because the energy of the protons is much greater than their rest mass, we have KE  E  pc .
A relationship for the magnetic field is given in section 32-1.
qBr
E
v
 mv  qBr  p  qBr 
 qBr 
m
c
 7.0 10 eV 1.60 10 J eV 
B

 5.5 T
qrc 1.60  10 C  4.25  10 m  3.00  10 m s 
19
15
E
19
3
8