Question

A radioactive substance decays 25%
of initial in 16 days. Its half life
period is:
(A) 8 days
(B) 28 days
(C) 32 days​

Answer 1

answer : option (A) 8 days.

explanation : Let initial amount of radioactive substance is $N_0$.

after 16 days , the radioactive substance reduces to 25% of its initial in 16 days.

so, remaining radioactive substance , N = 25 % of $N_0$

= $\frac{N_0}{4}$

using formula, $N=N_0e^{-\lambda t}$

or, $\frac{N_0}{4}=N_0e^{-\lambda 16}$

or, $\lambda=\frac{(ln4)}{16}$

now, half life of radioactive decay, $T_{1/2}=\frac{ln2}{\lambda}$

= $\frac{ln2}{\frac{ln(4)}{16}}$

= $\frac{16ln2}{ln2^2}$

= 16 × ln2/2ln2

= 16/2 = 8 days

hence, half life of radioactive substance is 8 days .