Solutions: Radioactivity Questions


(a) A radioactive source gives an initial count rate of 110 counts per second. After 10 minutes the count rate is 84 counts per second.

(background radiation = 3 counts per second)

(i) Give three origins of the radiation that contributes to this background radiation.

Any three of the following origins of background radiation gets two marks:

Two gets only one mark:

    • cosmic rays
    • ground, rocks and buildings
    • air
    • nuclear weapons testing/nuclear accidents
    • nuclear power discharge/waste from nuclear power stations
    • medical waste

(ii) Calculate the decay constant of the radioactive source in s–1.

Corrected original count of activity A0= 110 - 3 = 107 s–1

Corrected count of activity A after 10 minutes = 84 - 3 = 81 s–1

Time interval t = 10 minutes = 10 x 60 = 600s

ln (81/107) = - x 600

= -ln (81/107) /600 = 4.64 x 10-4

(use this value for further calculation - but quote the answer to this section to the correct number of sig figs)

= 4.6 x 10-4 s-1


(iii) Calculate the number of radioactive nuclei in the initial sample assuming that the detector counts all the radiation emitted from the source.

Initial decay rate = N0/t = 107Bq

N =107/ 4.64 × 10–4

= 2.3 × 105 radioactive nuclei

(7 marks)

(b) Discuss the dangers of exposing the human body to a source of radiation. In particular compare the dangers when the source is held outside, but in contact with the body, with those when the source is placed inside the body.

Alpha radiation is highly ionising, it therefore causes highly localised damage to cells and/or DNA. Damage to DNA can result in mutated cells or cancer. Cell damage can result in cell death.

If the alpha source is outside the body and in contact with the skin, emitted alpha particles will be emitted in all directions therefore half will be directed away from the body and ionise the air. Those directed towards the body will be absorbed by dead skin cells on the skin surface. They would therefore not result in damaged living tissue or cell mutation. An alpha source would only cause skin damage if it had a very high activity, then some skin burns would occur.

If the alpha source was inside the body it would be much more dangerous. All of the emitted alpha particles would be absorbed by living body tissue and cause possible cell mutations in a highly localised area. This would probably result in cancerous growth. If the alpha source was near to vital organs mutations would be even more dangerous as cancerous grwoths in such organs can easily lead to death. If the activity of the source was high, internal tissue damage (burns) would occur, if this was within an important vital organ the result would be very hazarous.

(Max 3 marks)

(Total 10 marks)