JEE Advanced Model Paper 3 with solutions for online practice

Let N0 be initial number of nuclei at time t = 0 .
The number of undecayed nuclei in time t is N = N0e−λt
The number of nuclei decayed in time t is
n = N0 - N = N0 - N0e−λt (i)
The number of undecayed nuclei in next time t,
N' = Ne−λt = (N0e−λt)e−λt = N0e−2λt (ii)
Number of decayed nuclei in next time t,
0.75 n = N - N' = N0e−λt - N0e−2λt (iii)
= N0e−λt(1−e−λt)
Dividing Eq. (ii) by Eq. (i), we get 0.75 = e−λt ⇒

4

3

= eλt
Taking natural logarithm , ln

4

3

= λt ⇒ λ =

ln 4 3

t

Given t = 2s ∴ λ =

ln4−ln3

2

=

2ln2−ln3

2

=

2×0.6931−1.0986

1

Mean life , τ =

1

λ

=

1

0.1438

= 7s (whole number)