The activity of a radioactive sample is measured as N_0 counts per minute at t=0 and N_0/e counts per minute at t=5 minutes. The time (in minutes) at which the activity reduces to half its value is

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Accepted Answer

Correct answer: D. (d) According to activity law R=R_0e^- λt ⋯ (i) According to given problem, R_0=N_0 counts per minute R= N_0e counts per minute t=5 minutes Substituting these values in equation (i), we get N_0e=N_0e^-5 e^-1=e^-5 5 =1 or = 15 per minute At t=T_1/2, the activity R reduces to R_02 where T_1/2= half life of a radioactive sample From equation (i), we get R_02=R_0e^- T_1/2 or e^ T_1/2=2 Taking natural logarithms of both sides of above equation, we get T_1/2= log_e2 or T_1/2= log_e2 = log_e2 ( 15 )=5 log_e2 minutes

$The activity of a radioactive sample is measured as N_{0} counts per minute at t = 0 and N_{0} / e counts per minute at t = 5 minutes. The time (in minutes) at which the activity reduces to half its value is$

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