The graph of the activity A of a radioactive sample against time t is shown below.
A / MBq
4
3
2
1
t/s
0
10
20
30
40
50
Find
(a) the decay constant of the sample.
(b) the number of undecayed nuclei at t = 0.
(c) the number of undecayed nuclei at t = 75.
(3 marks)
(2 marks)
(2 marks)
Answer:
(a) From the graph, the half-life of the sample is 10 s.
1
Applying k 
, the decay constant of the sample is
T 12
1
 0.1 s 1
10
(b) Applying A  kN , the number of undecayed nuclei at t = 0 is
A
4  10 6
N0  0 
 4  10 7
k
0.1
(c) Applying N  N 0 e  kt , the number of undecayed nuclei at t = 75 is
k


N  4  10 7  e 0.175  22120  22100
(1M)
(1M)
(1A)
(1M)
(1A)
(1M)
(1A)
*Code: 31L1B022, Total marks: 7
Xenon-133 (Xe-133) is a radioisotope used in lung and liver imaging. It has a half-life
of 126 hours and an atomic number of 54. Xenon-133 would undergo beta decay and
turn into caesium-133 (Cs-133).
(a) How many nucleons are there in a xenon-133 nucleus?
(1 mark)
(b) Write down the nuclear equation for the beta decay of xenon-133.
(2 marks)
(c) Find the decay constant of xenon-133 in s−1.
(2 marks)
(d) A sample of xenon-133 has an activity of 2 × 108 Bq. Calculate its activity after
240 hours.
(2 marks)
Answer:
(a) Xenon-133 has 133 nucleons.
133
0
(b) 133
54 Xe  55 Cs 1 e
(c) The decay constant of xenon-133 is
ln 2 ln 2
k

 5.501  10 3 h 1  1.528  10 6 s 1  1.53  10 6 s 1
T12 126
(d) The activity of the sample after 240 hours is
3
A  A0 e kt  2  108  e  5.50110 240  5.341  10 7  5.34  10 7 Bq


(1A)
(1A+1A)
(1M+1A)
(1M+1A)