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An Example Question

Background radiation count is found to be 20 s-1
A radioactive sample with an activity of 5 x 1011 Bq registers a count rate of 50 s-1.
Ten days later the count rate has dropped to 26 s-1.
(a) What is the new activity of the sample ?
(b) What is the half-life of the sample?
(c) How many radioactive nuclei were present at the initial measurement?

(a)

5 x 1011 Bq registers a count rate of 50 s-1
therefore 1 x 1010 Bq registers a count rate of 1 s-1
and 2.6 x 1011 Bq registers a count rate of 26 s-1

(b)

A/A0 = 6/30 and time interval t = 10 days

If A/A0 = ½ (=1/21) then t = 1 half-life
If A/A0 = ¼ (=1/22) then t = 2 half-lives
If A/A0 = 1/8 (=1/23) then t = 3 half-lives
Therefore if A/A0 = 1/2N then t = N half-lives

6/30 = 1/2N
1/5 = 1/2N
5 = 2N

Taking logs we get:

log10 5 = log10 2 x N
therefore N = 2.32

This means that 10 days = 2.32 half-lives, therefore one half-life is 4.3 days

Try this out on the excel worksheet

(c)

Now the initial activity was 5 x 1011 Bq
Half life was 4.3 days = 4.3 x 24 x 60 x 60 seconds = 3.7 x 105 s
Use the excel worksheet to check your maths