Cambridge Physics for the IB Diploma
Answers to Coursebook questions – Chapter 6.6
1
The distance of closest approach has been changed to 8.5 1015 m .
By conservation of energy:
1 2 k (2e)(79e)
2k (2e)(79e)
mv 
v
2
d
md
2  9 109 (2 1.6 10 19 )(79 1.6 10 19 )
v
6.4 1027  8.5 1015
v  3.7 107 m s 1.
2
The idea is that since the nucleus is very massive it will not recoil.
Then at the point of closest separation the kinetic energy will be a minimum and will
increase as the separation increases.
kZe 2
The potential energy is given by EP 
and so will be a maximum at the point of
r
closet separation and will tend to zero as the separation increases.
These observations give the graphs in the answers in the textbook (see page 808 in
Physics for the IB Diploma).
3
a
b
c
4
The velocity selector ensures that ions of the given speed continue undeflected.
This means that the electric and magnetic forces on the ions cancel out.
Hence qE  qvB  E  vB  2.0 105  0.15  3.0 104 N C1 .
mv 2
mv
From qvB 
we deduce that r 
.
qB
r
We are given the molar mass and, since one mole contains Avogadro’s number of
19.992 103
 3.3 1026 kg .
molecules, we have that the mass is
6.02 1023
3.3 1026  2.0 105
 0.083 m .
Hence, r 
1.6 1019  0.50
The radius for neon-22 will be r 
0.083  21.99
 0.091 m .
19.992
They are electrically neutral and so do not suffer scatterings the way a charged particle
would.
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5
Conservation of momentum gives: mv  mu  Mw
1 2 1 2 1
2
and conservation of energy gives: mv  mu  Mw .
2
2
2
mv  Mw
Mw
v
From the first equation, u 
and so
m
m
2
1 2 1 
Mw  1
2
mv  m  v 
  Mw .
2
2 
m  2
Solving for the nucleus speed we get:
1 2 1 2
1 M 2 w2 1
mv  mv  Mwv 
 Mw2
2
2
2 m
2
2 2
2
0  2Mmwv  M w  Mmw
0  2mwv  ( M  m) w2
This means that either w  0 , which is not realistic, or w 
6
a
We know that  
b
We start with
2mv
as required.
mM
ln 2 ln 2

 0.231 s 1 .
T1/2 3.00
1
 6.02 1023  6.02 1021 nuclei and so:
100
i
N  6.02 1021  e0.2311  4.78 1021
ii
N  6.02 1021  e0.2312  3.79 1021
iii
N  6.02 1021  e0.2313  3.011021
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7
1
.
2
a
The probability of decay within a half-life is always
b
The probability that the nucleus will not decay after the passage of three half1
1
1
1
lives is


 .
2
2
2
8
no decay
no decay
no decay
Hence the probability that the nucleus will decay some time within three half1 7
lives is 1    0.875 .
8 8
c
The probability of decay in any one half-life interval is 0.5.
More mathematically, we want to find P(D N ) , where we use the notation of
conditional probability, and the events D and N stand for D = decay in the next
half-life and N = no decay in the first 4 half-lives, respectively.
P(D  N )
Then P(D N ) 
.
P(N )
1
1
1
1
Now, P(N )  4 
and P(D  N ) 
. Hence P(D N )  .
2
16
32
2
8
The half-life is so long, so that what we are really asked to find is the initial activity of
1.0 g of pure radium.
We have that A   N 0 e t , so that the initial activity is  N 0 .
1.0
 0.0044 moles
A mass of 1.0 g of radium corresponds to
226.005
and hence N 0  0.0044  6.02 1023  2.65 10 21 nuclei.
ln 2
ln 2

 1.37 1011 s 1
Since  
T1/2 1600  365  24  60  60
we find an activity of 1.37 1011  2.65 1021  3.6 1010 Bq .
9
The decay constant is  
ln 2 ln 2

 0.0578 d 1
T1/2 12
and so A   N 0 e t  3.5  e 0.057820  1.1 MBq .
10
The decay constant is  
ln 2
ln 2

 1.34  106 s 1 .
T1/2 6  24  60  60
From A   N 0 e t we find
0.50 106  1.34 106  N0e1.3410
6
246060
 N0  4.2 1011 .
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11
After time t the number of uranium atoms remaining in the rocks is N  N 0 e t
and so the number that decayed (and hence eventually became lead) is
N  N 0  N 0 (1  e  t ) .
N 0 (1  e t )
 0.80 .
Hence we have that
N 0 e  t
This means that 1  e t  0.80e  t  1  1.80e  t  et  1.80 .
Hence t  ln(1.80)  0.5878 .
0.5878
ln 2
ln 2
 3.8 109 y .

 1.54 1010 y 1 we find t 
Since  
10
9
1.54 10
T1/2 4.5  10
12
The method of Q11 may be used but here, clearly, a ratio of 1 to 7 corresponds to three
half-lives and so the age is about t  3 1.37 109  4.1109 y .
13
a
In any experiment the detector will receive radiation from natural sources (the
‘background’) in addition to the radiation from the radioactive source used in the
experiment. The background is always present and so it is registered even after
the activity of the source goes to zero.
b
The background activity of about 20 Bq has been subtracted from each of the
measurements. From the second graph we obtain a half-life of just over 7
minutes.
14
The activity is given by A   N   N 0 e  t where  
a
AA  A N 0 A 3
3

 1 .
AB  B N 0 B 4
4
b
AA A N 0 Ae  A 4 3 e 4

 
 0.95 .
AB B N 0 B e  B 4 4  ln32 4
e
c
AA A N 0 Ae  A 12 3 e 4

 
 1.5 .
AB B N 0 B e  B 12 4  ln32 12
e

ln 2

ln 2
is the decay constant.
T1/2
4
ln 2
12
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15
This is a very difficult question and many different possibilities must be
considered. Essentially we must be able to determine from a graph of activity
versus time the initial activities of the two isotopes and their respective halflives. One possibility is represented by the following graph in which a short (S) and a
long (L) half-life isotopes are present. The shape of the curve is not a pure exponential.
A x10 ^10 Bq
2
1.5
1
0.5
t min
1
2
3
4
We see that after about 1 minute we have a smooth exponential curve which implies
that one of the isotopes has essentially decayed away, leaving behind just one isotope.
This is justified by estimating a half-life for times greater than 1 minute. We get
consistently a half- life of 1 minute for the long half- life isotope. Extending the
exponential curve smoothly backwards, we intercept the vertical axis at about
11010 Bq .
A x10 ^10 Bq
2
1.5
1
0.5
t min
1
2
3
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4
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Thus the activity of isotope L is given by AL  1010  0.5t /1 . This means that the
initial activity of the other isotope is also 11010 Bq . Subtracting from the data
points of the given graph the activity of this isotope we get the following graph:
A x10 ^10 Bq
1
0.8
0.6
0.4
0.2
t min
0.2
0.4
0.6
0.8
1
This represents the decay of just isotope S. From this graph we find a half-life of about
0.1 minute.
Obviously, this analysis gets more complicated when the half-lives are not so different
or when the initial activities are very different.
16
a
If the mass (in grams) is m and the molar mass is  , the number of moles of the
m
radioactive isotope is
.

The initial number of nuclei is then N 0 
m

 N A since one mole contains
Avogadro’s number of molecules.
b
The activity is A   N   N 0 e t  
A0  
m

m

N Ae t and the initial activity is thus
N A . Measuring the initial activity then allows determination of the
decay constant and hence the half-life from  
17
ln 2
.
T1/2
The radius of a nucleus of mass number A is R  1.2  A1/ 3  10 15 m and its mass is
M  Amn (here m n is the mass of a nucleon).
A  mn
mn
M


The density is therefore  
4 3 4
4
πR
π(1.2  A1/3 1015 )3
π(1.2 10 15 )3
3
3
3
and so is independent of A.
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18
a
The force becomes negligible at separations of about 1.5 1015 m .
b
From the graph this is about 4 kN.
c
F
9 109  1.6 1019 
1.0 10 
15 2
2
 230 N .
6.67 1011  1.67 1027 
d
F
e
Fe
230

 1036 .
34
Fg 2 10
1.0 10 
15 2
2
 2 1034 N .
19
Generally, this process takes place in heavy atoms and the captured electron is taken
from the inner shell of electrons in the atom. This means that there will follow a
transition of an atomic electron down to the ground state with the emission of a
(typically X-ray) photon. Another way of telling is that the atom after the electron
capture has changed atomic number, i.e. it is a different element.
20
a
b
226
88
Ra 
226
88
Ra  00 γ
hc
hc
 E   
.

E
6.63 1034  3.0 108
 1.83 1011 m .
Hence  
0.0678 106 1.6 1019
hf 
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