Question

At time t=0, activity of a radioactive substance is 1600 Bq, at t=8 s activity remains 100 Bq. Find the activity at t=2 s.

Answer 1

Activity after 2sec will be 800Bq •) Now we know that at t=0 activity is N0 , where N0 is the initial activity then after time t the activity is given as N = N0 × (1/2)^n , where N is the activity at some time.

where, n = t / t½ , where t½ is the half life of the substance.

•)For given case, N = 100Bq , N0 = 1600Bq , now ,

N = N0 (1/2)^n

100 = 1600 × (1/2) ^n

=> ( 1/2)^n = 1 / 16

=> n = 4

•) Now, n = t / t½

=> t½ = 8 /4 = 2sec

So half life period of the substance is 2 seconds , hence, activity after 2 seconds will be 1600/2=800Bq