1.The activity of a radioactive sample is measured as N0N_0N0 counts per minute at t=0t = 0t=0 and N0e\frac{N_0}{e}eN0 counts per minute at t=5t = 5t=5 min. The time (in minutes) at which the activity reduces to half its value isa.loge25\log_e \frac{2}{5}loge52b.5loge2\frac{5}{\log_e 2}loge25c.5log1025 \log_{10} 25log102d.5loge25 \log_e 25loge2Login to continueOnly logged in users canattempt or see the solution.