Question
The count rate of a Geiger Muller counter for the radiation of a radioactive material of half-life $$30\,\min $$ decreases to $$5\,{s^{ - 1}}$$ after $$2\,h.$$ The initial count rate was
A.
$$20\,{s^{ - 1}}$$
B.
$$25\,{s^{ - 1}}$$
C.
$$80\,{s^{ - 1}}$$
D.
$$625\,{s^{ - 1}}$$
Answer :
$$80\,{s^{ - 1}}$$
Solution :
The equation for initial and final count rate is
$$\eqalign{
& N = {N_0}{\left( {\frac{1}{2}} \right)^n}\,\,\left[ {_{{N_0} = {\text{ initial rate of radio - active atom}}}^{N = \,\,{\text{final count rate}}}} \right] \cr
& {\text{where,}}\,n = \frac{t}{{{T_{\frac{1}{2}}}}} \cr
& {\text{Here,}}\,\,n = \frac{{120}}{{30}} = 4\,\,\left[ {\because t = 2h = 2 \times 60\min = 120\min } \right] \cr
& \therefore \frac{N}{{{N_0}}} = {\left( {\frac{1}{2}} \right)^4} = \frac{1}{{16}} \cr
& {\text{or}}\,\,{N_0} = 16 \times N = 16 \times 5 = 80\,{s^{ - 1}} \cr} $$