Q#22
(a) A radiation detector is placed close to a radioactive source. The detector does not surround the source.
Radiation is emitted in all directions and, as a result, the activity of the source and the measured count rate are different.
Suggest two other reasons why the activity and the measured count rate may be different. [2]
(b) The variation with time t of the measured count rate in (a) is shown in Fig. 12.1.
Fig. 12.1
(i) State the feature of Fig. 12.1 that indicates the random nature of radioactive decay. [1]
(ii) Use Fig. 12.1 to determine the half-life of the radioactive isotope in the source. [4]
(c) The readings in (b) were obtained at room temperature.
A second sample of this isotope is heated to a temperature of 500 °C.
The initial count rate at time t = 0 is the same as that in (b).
The variation with time t of the measured count rate from the heated source is determined.
State, with a reason, the difference, if any, in
1. the half-life,
2. the measured count rate for any specific time.
[3]
[Total: 10]
Solution:
(a)
Any two points
emission from radioactive daughter products
self-absorption in source
absorption in air before reaching detector
detector not sensitive to all radiations
window of detector may absorb some radiation
dead-time of counter
background radiation
(b)
(i)
The curve is not smooth.
(ii)
clear evidence of allowance for background
half-life determined at least twice
half-life = 1.5 hours
{As the time increases, the curve should tend towards zero. However, it is observed from the graph, that the curve tends towards a count rate of 10. So,
Background radiation = 10 counts / min
This value should be reduced from the count rate at different times.
In 1 half-life, the count rate should be halved (after accounting for the background radiation).
1st set of data:
From graph, when count rate = 160, time = 0.2 hour
This corresponds for a count rate of (160 – 10 =) 150 for the radioactive isotope only.
After 1 half-life, this value would be halved (= 150 / 2 = 75). In the graph (after adding the background radiation), the count rate would be (75 + 10 =) 85.
From graph, when count rate = 85, time = 1.7 hour
Half-life = 1.7 – 0.2 = 1.5 hour
2nd set of data:
From graph, when count rate = 100, time = 1.3 hour
This corresponds for a count rate of (100 – 10 =) 90 for the radioactive isotope only.
After 1 half-life, this value would be halved (= 90 / 2 = 45). In the graph (after adding the background radiation), the count rate would be (45 + 10 =) 55.
From graph, when count rate = 55, time = 2.8 hour
Half-life = 2.8 – 1.3 = 1.5 hour
Averaging the 2 values give a half-life of 1.5 hour.}
(c)
1. There is no change in the half-life because the decay is spontaneous/independent of environment
2. The count rate (is likely to be or could be) different
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