A radioactive sample at any instant has its disintegration rate 5000 disintegrations per minute After 5 minutes, the rate is 1250 disintegration per minute. Then, find out the decay constant (per minute)-
(A) $0.4 ln2$
(B) $0.2 ln2$
(C) $0.1 ln2$
(D) $0.8 ln2$

Hint
The decay constant (symbol: $λ$ and units: $s^{−1}$ or $a^{−1}$) of a radioactive nuclide is its probability of decay per unit time. The decay constant relates to the half-life of the nuclide $T_{½}$ through $T_{½} = {ln 2}/λ$.

Complete step by step answer
We know, Number of nuclide is $N= {N_0}e^{-\lambda t}$ … (1)
We also know that activity $A$ is directly proportional to $N$-
$A = \dfrac{{dN}}{{dt}} \times N$
So, we can replace N with A
So, continuing equation 1
$A = {A_0}{e^{ - \lambda t}}$ ..........equation 2
${e^{\lambda t}} = \dfrac{{{A_0}}}{A}$
Taking log in both sides,
$\lambda = \dfrac{1}{t}\ln (\dfrac{{{A_0}}}{A})$
Now put the values which are $t = 5$, $A_0 = 5000, A = 1250$.
$ \Rightarrow \lambda = \dfrac{1}{5}\ln (\dfrac{{5000}}{{1250}})$
$
   \Rightarrow \lambda = 0.2\ln 4 \\
   \Rightarrow \lambda = 0.4\ln 2 \\
 $
So, the decay constant is $0.4 ln2$. Option (A) is correct.

Note
An unstable nucleus spontaneously emits particles and energy in a process known as radioactive decay. The term radioactivity refers to the particles emitted. When enough particles and energy have been emitted to create a new, stable nucleus (often the nucleus of an entirely different element), radioactivity ceases.