Question

A radioactive isotope has a half life of T years. How long will it take the activity to reduce to a) 3.125%, b) 1% of its original value?

Answer 1

The fraction of the original sample is left after 5 years and also the fraction of the original sample is left after 6.64 years.

Explanation:

(a)

The fraction of the original sample left is

N/No = 3.125 / 100

         = 1/32 = (1/2)^5

Hence there are 5 half lives of T years spent. Thus the time taken is 5T years.

(b)

The fraction of the original sample left = 1/100 = (1/2)^n

i.e. 2^n = 100

n = log 100 / log 2 = 2 / 0.301  = 6.64 T

Thus we have t =  6.64 T