Question

At time t=0 , number of nuclei of a radioactive substance are 100. At t=1 s these numbers become 90. Find the number of nuclei at t=2 s.

Answer 1

At t = 2s No. of active nucleus will be 81

•According to Radioactivity decay law rate of decay of nucleus is directly proportional to no. of undecayed nucleus left .

•i.e. so, N = N•e^-kt

where N is Undecayed nucleus left

N• is No. of nucleus in starting

k = decay constant

t = time spend

•

N• = 100

At t = 1s ,

N = 90

90 = N•e^-kt

90 = 100(e^-k)

(e^-k) = 90/100 = 9/10

• we get value of (e^-k) = 9/10 ___(1)

•At t = 2s

N = N•e^-kt

N = 100 ( e^-2k)

N = 100 ( e^-k)² _____(2)

•On putting (e^-k) = 9/10 in (2)

we get,

N = 100 (9/10)²

N = (100 × 81)/100

N = 81

•At t = 2s No. of active nucleus will be 81