(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?

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

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 = 1/2^N then t = N half-lives

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

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

(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