Names:________________________________________________________________________________
Physics 103 Team Session for Chapter 31 & 32
1) In electrically neutral atoms, how many
(a) protons are in the uranium 238
92 U nucleus? (b) neutrons are in the mercury
(c) electrons are surrounding the
93
41 Nb
202
80 Hg nucleus?
nucleus?
REASONING
a.
The number of protons in a given nucleus
A
ZX
b.
The number N of neutrons in a given nucleus
of protons and neutrons) minus the atomic
N = A  Z (Equation 31.1).
is specified by its atomic number Z.
A
ZX
is equal to the nucleon number A (the number
number Z (the number of protons):
c.
In an electrically neutral niobium atom, the number of electrons in orbit about the nucleus is
equal to the number of protons in the nucleus.
SOLUTION
a.
The number of protons in the uranium
b. The number N of neutrons in the
202
80 Hg
c. The number of electrons that orbit the
238
92 U nucleus
is Z = 92 .
nucleus is N = A  Z = 202  80 = 122 .
93
41 Nb nucleus
in the neutral niobium atom is equal to the number of
protons in the nucleus, or 41 .
2) What is the radius of a nucleus of titanium
48
22 Ti ?
REASONING AND SOLUTION Equation 31.2 gives for the radius of the
48
22 Ti
nucleus that
r = (1.2  10–15 m)A1/3 = (1.2  10–15 m)(48)1/3 = 4.4  10 15 m
(3) Iodine
131
53 I
is used in diagnostic and therapeutic techniques when treating thyroid disorders. The isotope
has a half-life of 8.04 days. What percentage of the initial sample remains after 30.0 days?
REASONING AND SOLUTION According to Equation 31.5, the fraction of an initial sample remaining after
a time t is N / N0  e –  t , where  is the decay constant. The decay constant is related to the half-life T1/ 2 .
According to Equation 31.6, the decay constant is   0.693/ T1/ 2 . Therefore, the fraction remaining is
N
–0.693 30.0 days / 8.04 days 
 e –0.693t /T1/2  e
 0.0753
N0
This fraction corresponds to a percentage of 7.53 % .
4) Determine the mass defect (in atomic mass units) for
(a) helium 23 He , which has an atomic mass of 3.016 030 u and
(b) the isotope of hydrogen known as tritium 31 T , which has an atomic mass of 3.016 050 u.
(c) On the basis of your answers, which nucleus requires more energy to disassemble?
(mp = 1.007 825 u and mn = 1.008 665 u )
REASONING The mass defect is the total mass of the stationary separated nucleons (protons and neutrons)
minus the mass of the intact nucleus. The given atomic masses are for the electrical neutral atoms and,
therefore, include the mass of the electrons. This will cause no problem, provided that we use the atomic
mass of the hydrogen atom (including its electron) when determining the mass of the separated nucleons.
Referring to Table 31.1 we find that the mass of a neutron is 1.008 665 u and the mass of a hydrogen atom
is 1.007 825 u.
SOLUTION
a. The helium 23 He nucleus contains 2 protons and 3  2 = 1 neutron. Thus, the mass defect m is
 m  2 1.007 825 u   11.008 665 u   3.016 030 u  0.008 285 u
Separated nucleons: 2 protons and 1 neutron
Intact nucleus
b. The tritium 31 T nucleus contains 1 proton and 3  1 = 2 neutrons. Thus, the mass defect m is
 m  11.007 825 u   2 1.008 665 u   3.016 050 u  0.009 105 u
Separated nucleons: 1 proton and 2 neutrons
Intact nucleus
c. The mass defect for tritium 31 T is greater than that for helium 23 He . The mass defect is related to the
binding energy as follows:
Binding energy   m  c2
(31.3)
The binding energy is the energy that must be supplied to the intact nucleus in order to separate it into its
constituent nucleons. Thus,
more energy must be supplied to tritium 31 T than to helium 23 He .
5) The biological equivalent dose for a typical chest X-ray is 2.5 x 10-2 rem. During a particular chest Xray, the mass of exposed tissue was 21 kg and it absorbed 6.2 x 10-3 J of energy. What is the relative
biological effectiveness (RBE) for the radiation on this particular type of tissue?
REASONING The relation between the rad and gray units is presented in Section 32.1 as 1 rad = 0.01 gray. If,
for instance, we wanted to convert an absorbed dose of 2.5 grays into rads, we would use the conversion
procedure:
 1 rad 
  250 rad
 0.01 Gy 
 2.5 Gy  
In general, the conversion relation is
 1 rad 
Absorbed dose  in rad    Absorbed dose  in Gy  

 0.01 Gy 
(1)
SOLUTION
According to Equation 32.4, the relative biological effectiveness (RBE) is given by
RBE =
Biologically equivalent dose
Absorbed dose (in rad)
The absorbed dose (in rad) is related to the absorbed dose (in Gy) by Equation (1), so the RBE can be expressed
as
RBE =
Biologically equivalent dose
 1 rad 
 Absorbed dose  in Gy   

 0.01 Gy 
The absorbed dose (in Gy) is equal to the energy E absorbed by the tissue divided by its mass m
(Equation 32.2), so the RBE can be written as
Biologically equivalent dose
2.5  102 rem
RBE =

 0.85
 6.2  103 J   1 rad 
 E   1 rad 


 
 

 m   0.01 Gy 
 21 kg   0.01 Gy 