Question
The half-life of a radioactive substance is 30 minutes. The time (in minutes) taken between $$40\% $$ decay and $$85\% $$ decay of the same radioactive substance is
A.
15
B.
30
C.
45
D.
60
Answer :
60
Solution :
Key Idea
Half-life of a radioactive substance is $${T_{\frac{1}{2}}} \propto \log \left( {\frac{{{N_0}}}{N}} \right)$$
Given, $${N_1} = 0.6\,{N_0}\,\,\left( {\because 40\% \,{\text{decay}}} \right)$$
$${N_2} = 0.15\,{N_0}\,\,\left( {\because 85\% \,{\text{decay}}} \right)$$
Putting these in the formula,
$$\frac{{{N_2}}}{{{N_1}}} = \frac{{0.15\,{N_0}}}{{0.6\,{N_0}}} = \frac{1}{4} = {\left( {\frac{1}{4}} \right)^2}$$
So, two half-life periods has passed.
Thus, time taken $$ = 2 \times {t_{\frac{1}{2}}} = 2 \times 30 = 60\,\min $$