Question

The rate of disintegration of a radioactive substance falls from $800decay/min $ to $100{\text{ }}decay/min$ in $6 hours$. The half-life of the radioactive substance is:
(A) $6/7\,hr$
(B) $2 hr$
(C) $3 hr$
(D) $1 hr$

Answer 1

Hint
According to the law of radioactive disintegration $N = {N_0}{e^{ - \lambda t}}$. Substitute the data and calculate the radioactive constant. Then we will calculate the half life of the element by the relation-
${T_{1/2}} = \dfrac{ln2}{\lambda}$.

Complete step-by-step solution
At any instant the rate of decay of radioactive atoms is proportional to the number of atoms present at that instant. It is given by,
 $N = {N_0}{e^{ - \lambda t}}$
Where, $N$is the number of atoms remaining undecayed after time $t$,
 ${N_0}$ Is the number of atoms present initially,
 $\lambda $ Is the decay constant.
Given that,
 $N = 100\,\,decay/\min $
 ${N_0} = 800\,\,decay/\min $
 $t = 6\,\,hr$
Substitute the data in the expression.
 $100 = 800{e^{ - \lambda (6 \times 60)}}$
 ${e^{ - 360\lambda }} = \dfrac{1}{8}$
 $360\lambda = \ln 8$
 $\lambda = \dfrac{{\ln {2^3}}}{{360}}$
 $\lambda = \dfrac{{\ln 2}}{{120}}$
Half life of the element is given by,
 ${T_{1/2}} = \dfrac{ln2}{\lambda}$.
Substitute the value of decay constant.
${T_{1/2}} = \dfrac{ln2}{{ln2}/{120}}$
${T_{1/2}} = 120 min = 2 hrs$
Hence, the half life of the radioactive substance is ${T_{1/2}} = 2 hrs$
The correct option is (B).

Note
Half life is the time interval in which mass of a radioactive substance or the number of its atom to reduce to half of its initial value.
Activity is defined as the rate of disintegration of the substance. It is given by,
 $A = - \dfrac{{dN}}{{dt}}$
The unit of activity is Becquerel, Curie and Rutherford.