Question

A radioactive sample at any instant has its disintegration rate $5000$ disintegration per minute. After $5$ minutes, the rate is $1250$ disintegrations per minute. Then, the decay constant, per minute, is

Answer 1

Let decay constant per minute$=\lambda$

Therefore, ATQ, initially disintegration rate,

$R=N_{\circ }\lambda =5000\to \left(1\right)$

Also, in 2nd case, disintegration rate after 5 min is,

$1250=N\lambda\rightarrow (2)$

Dividing equation (1) by (2), we get

$\frac{N_{\circ }\lambda }{N\lambda }=\frac{6000}{1250}\frac{N_{\circ }}{N}=4$

And as we know, the number of decay, N is given by,

$N=N_{\circ }{e}^{-\lambda t}(t=5min)\frac{1}{4}=e^{-5\lambda }-ln\left(4\right)=-5\lambda \lambda =0.4ln2$