Question

if the activity of a radioactive sample drops to 1/16 of its initial value in3 hours,It's half life is​

Answer 1

using radioactive decay equation,

$N=N_0e^{-\lambda t}$

here it is given that a radioactive same drops to 1/16 of its initial value in 3 hours.

i.e., $N=\frac{N_0}{16}$, t = 3hours

so, $\frac{N_0}{16}=N_0e^{-\lambda 3}$

or, $\frac{1}{16}=e^{-3\lambda}$

or, $2^{-4}=e^{-3\lambda}$

taking log base e both sides,

or, $-4log2=-3\lambda$

or, $\frac{4}{3}log2=\lambda$....(1)

half life of radioactive substance is given by, $T_{1/2}=\frac{log2}{\lambda}$

From equation (1),

= $\frac{log2}{\frac{4}{3}log2}$

= $\frac{3}{4}$ hour.