Solutions: Radioactivity Questions
Q15.
(a) Calculate the radius of the 
nucleus.
r0 = 1.3 × 10–15 m
A = 238
R = 1.3 × 10–15 x 2380.333
R = 8.06 × 10–15m
R = 8.06 fm
(2 marks)
(b) At a distance of 30 mm from a point source of γ rays the corrected count rate is C.
Calculate the distance from the source at which the corrected count rate is 0.10 C, assuming that there is no absorption.
x12 I1 = x22 I2
x1 = 30 mm
x2 = ? mm
I1 = C
I2 = 0.10C
302C = x22 x 0.10C
x22 = 9000
x2 = 95 mm
(2 marks)
(c) The activity of a source of β particles falls to 85% of its initial value in 52 s.
Calculate the decay constant of the source.
N = 0.85 N0
N = N0e-λt
0.85 N0 = N0e-λt
0.85 = e-λt
Taking logs:
ln 0.85 = -λt
t = 52 s
λ = -(ln 0.85)/52
λ = 3.1 x 10-3 s-1
(3 marks)
(d) Explain why the isotope of technetium, Tc 99m, is often chosen as a suitable source of radiation for use in medical diagnosis.
Technicium 99m may be prepared on site within the hospital.
It only emits γ rays as the metastable nucleus moves to a stable state.
There is no chain of decay assocaited with it, therefore no residual daughter radioactive nuclei will remain active in the body after use.
As is only emits gamma rays that means that any ionisation of tissue that does occur will be widely spaced and the body's defense mechanisms will be better able to cope with any damage.
It is easy to detect the emission sites of the gamma rays using a gamma camera so that diagnosis can be made.
It has a short enough half-life to not make the patient radioactive for too long, and yet the half life is long enough for the medical tests to be able to be carried out.
The substance itself has a toxicity that can be tolerated by the body.
(3 marks maximum
(Total 10 marks)
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The intensity of the source directly affects the count rate so we can use the count rate as the intensity value.
