State the law of radioactivity and hence, show that `N=N_(0)e^(-lambda t)`.

Explanation

Statement : In any radioactive sample which undergoes `alpha, beta,gamma-` decay it is found that the number of nuclei undergoing the decay per unit time is proportional to the total number of nuclei in the sample. If N is the number of nuclei in the sample and `Delta` N undergo decay in the time `Delta`t then `(Delta N)/(Delta t) alpha Nimplies (Delta N)/(Delta t)=lambda N` Where `lambda` is called radioactive decay constant (or) disintegration constant The change in the number of nuclei in the sample is `dN=-Delta N` in time `Delta t` Thus the rate of change of N is (in the limit `Delta t to 0` ) `(dN)/(dt)=-lambda Nimplies (dN)/(N)=- lambda dt` Now integration on both sides `int_(N_9)^(N) (dN)/(N)=-lambda int_(t_0)^(l)dt " in" N-In N_(0)=-lambda(t-t_(0))` Here `N_(0)` is the number of radioactive nuclei in the sample at some arbitarary time `t_(0)` and N is the number of radioactive nuclei at any subsequent time t, but `t_(0)`=0, then In `((N)/(N_(0)))=-lambda t implies (N)/(N_(0))=e^(-lambda t)`
`N=N_(0)e^(-lambda t)`