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Physics 112
Homework 13 (solutions)
(2004 Fall)
Solutions to Homework Questions 13
Chapt29, Problem-2:
What is the order of magnitude of the number of protons in your body? Of
the number of neutrons? Of the number of electrons?
Solution:
Of course the body contains many different elements, and some of these are of high Z number (such
as iron) and so they contain a few more neutrons than protons. However by far the majority of the
atoms in your body are of low Z number – such as H, C, O etc. Indeed ~70% of your body is H20 in
which there are a total of eight neutrons and ten protons. So for this order-of-magnitude estimate,
we can assume the protons and neutrons are nearly equally numerous in your body, each contributing
(say) 35 kg out of a total body mass of 70 kg.
" 1 nucleon %
28
28
N = 35 kg $
-27
' ~10 protons and ~10 neutrons
# 1.67 !10 kg &
You are approximately neutral (uncharged) so electron number is precisely equal to the proton
number,
28
N e = ~10 electrons
Chapt29, Problem-3:
Using the result of Example 29.1, find the radius of a sphere of nuclear
matter that would have a mass equal to that of Earth. (Earth has a mass of 5.98x102 4 kg and equatorial radius
of 6.38x106 m.)
Solution:
&
#4
From the expression relating mass, volume & density M E = !n V = ! n % " r 3 ( , so we can trivially
$3
'
rearrange this to find the radius
13
* 3 5.98 ) 1024 kg
# 3 ME &
r=%
( = ,,
17
3
$ 4 ! "n '
+ 4 ! 2 .3 )10 kg m
(
(
)
)
/
/
.
13
2
= 1.8 ! 10 m
Chapt29, Problem-4:
Consider the hydrogen atom to be a sphere of radius equal to the Bohr radius,
0.53x10– 1 0 m, and calculate the approximate value of the ratio of the nuclear density to the atomic density.
Solution:
The mass of the hydrogen atom is approximately equal to that of the proton, 1.67 !10 "27 kg . If the
radius of the atom is r = 0.53 ! 10"10 m , then
3 1.67 # 10$27 kg
m
m
!a = =
=
= 2.7 # 103 kg m 3
V ( 4 3 ) " r 3 4" 0.53 # 10$10 m 3
(
(
)
)
The ratio of the nuclear density to this atomic density is
!n 2 .3 "1017 kg m3
8.6 !1013
=
3
3 =
!a 2.7 "10 kg m
1
Physics 112
Homework 13 (solutions)
Chapt29, Problem-12:
(2004 Fall)
56
The peak of the stability curve occurs at Fe. Elements up to iron are
produced in the cores of massive stars by exothermic fusion reactions. This is the fundamental reason that
iron and lighter elements are much more common in the Universe than elements with higher mass numbers.
Show that 5 6Fe has a higher binding energy per nucleon than its neighbors 5 5Mn and 5 9Co. Compare your
results with Figure 29.4.
Solution:
The difference in mass is !m = ZmH + ( A " Z ) mn " m and the binding energy (per nucleon) is
therefore given by Eb A = !m ( 931.5 MeV u ) A so we can construct the following table
Nucleus
55
25 Mn
Z
25
(A – Z )
30
m ( in u )
!m ( in u )
Eb A
54.938 048
0.517 527
8.765
56
26 Fe
59
27 Co
26
30
55.934 940
0.528 460
8.786
27
32
58.933 198
0.555 357
8.768
(in MeV )
It can be seen that
indeed has a greater binding energy per nucleon than its neighbors. (This
gives us finer detail than is shown in Figure 29.4.)
56
26 Fe
Chapt29, Problem-15:
The half-life of an isotope of phosphorus is 14 days. If a sample contains
3.0x101 6 such nuclei, determine its activity. Express your answer in curies
Solution:
The decay constant is ! =
ln 2
, so the activity is
T1 2
3.0 " 1016 ln 2
N ln 2
R = !N =
=
= 1.7 "1010 decays s ,
T1 2
(14 d ) 8.64 "104 s d
(
)
(
or
)
"
1 Ci
R = 1.7 !1010 decays s $
10
# 3.7 !10 decays
(
)
%
' = 0.46 Ci
s&
Chapt29, Problem-16:
A drug tagged with (half-life = 6.05 h) is prepared for a patient. If the
original activity of the sample was 1.1x104 Bq, what is its activity after it has sat on the shelf for 2.0 h?
Solution:
The activity is R = R0 e !"t where ! =
R = R0 e
(
! tln 2 T1 2
ln 2
. Thus,
T1 2
) = 1.1 "104 Bq e !
(
)
( 2.0 h) ln2
6.05 h
3
= 8.7 ! 10 Bq
2
Physics 112
Chapt29, Problem-22:
Homework 13 (solutions)
(2004 Fall)
14
After a plant or animal dies, its C content decreases with a half-life of 5
730 yr. If an archaeologist finds an ancient firepit containing partially consumed firewood, and the 1 4C
content of the wood is only 12.5% that of an equal carbon sample from a present-day tree, what is the age of
the ancient site?
Solution:
Using R = R0 e ! "t , with R R0 = 0.125 , gives !t = " ln ( R R0 )
or
t=!
ln ( R R0 )
"
# ln ( R R0 ) &
# ln ( 0.125 ) &
4
( = !( 5730 yr ) %
= ! T1 2 %
( = 1.72 ! 10 yr
%$ ln 2 ('
$ ln 2 '
Chapt29, Conceptual-1: Isotopes of a given element have many different properties, such as mass,
but the same chemical properties. Why is this?
Solution:
Isotopes of a given element correspond to nuclei with different numbers of neutrons. This results in
a variety of different properties for the nuclei, including the obvious one of mass. The chemical
behavior, however, is governed by the lectrons. All isotopes of a given element have the same
number of electrons and, therefore, the same chemical behavior.
Chapt29, Conceptual-3: A student claims that a heavy form of hydrogen decays by alpha emission.
How do you respond?
Solution:
An alpha particle contains two protons and two neutrons. However a hydrogen nucleus only contains
a single proton, so it cannot emit an alpha particle.
Chapt29, Conceptual-5: In beta decay, the energy of the electron or positron emitted from the
nucleus lies somewhere in a relatively large range of possibilities. In alpha decay, however, the alpha-particle
energy can only have discrete values. Why is there this difference?
Solution:
In alpha decay, there are only two final products – the alpha particle and the daughter nucleus.
There are also two conservation principles – energy and momentum. As a result, the alpha particle
must be ejected with a discrete energy to satisfy both conservation principles. However, beta decay
is a three-particle decay – the beta particle (electron or positron) the anti-neutrino (or a neutrino)
and the daughter nucleus. As a result, the energy and momentum can be shared in a variety of ways
between the three particles while still satisfying the two conservation principles. This allows a
continuous range of energies for the beta particle.
3
Physics 112
Homework 13 (solutions)
(2004 Fall)
Chapt29, Conceptual-8: An alpha particle has twice the charge of a beta particle. Why dooes the
former deflect less than the latter when passing between electrically charged plates, assuming they both have
the same speed?
Solution:
The much larger mass of the alpha particle as compared to that of the beta particle ensures that it
will not deflect as much as does the beta, which has a mass about 7000 times smaller.
Chapt29, Conceptual-9: Can carbon-14 dating be used to measure the age of a stone?
Solution:
Carbon dating cannot generally be used to estimate the age of a stone because the stone was not
alive to ‘take up’ carbon from the environment. Only the ages of artifacts that were once alive can
be estimated with carbon dating.
4