# Define the term ‘decay constant’ of a radioactive sample. The rate of disintegration of a given radioactive nucleus is 10000 disintegrations/s and 5,000 disintegrations/s after 20 hr. and 30 hr. respectively from start. Calculate the half-life and initial number of nuclei at t = 0.

Decay constant of a radioactive substance is the ratio of the rate of change of the number nuclei at any given time in the sample to the number of nuclei in the sample at that time :

Let the number of nuclei be N. and $\lambda$ be the decay constant.

$\frac{dN}{dt}= - \lambda N$

We know that,

$R = R_{o}e^{-\lambda t}$

On equating the given values, we get

$10000= R_{o}e^{-\lambda 20} \; \; \; \; \; -----(a)$

$5000= R_{o}e^{-\lambda 30} \; \; \; \; \; -----(b)$

On dividing both the equations (a) and (b) we have

$2= e^{10 \lambda}$

On taking in both sides, we have

$\lambda =\frac{In \; 2}{10}=0.0693\: hr^{-1}$

$\lambda = 1.925\times 10^{-5}\; s^{-1}$

The half-life is  -

$t_{1/2}=\frac{ln\; 2}{\lambda } = 10 hr$

Now, let's find $R_{o}$

$10000= R_{o}e^{-\lambda 20}$

$R_{o}=\frac{10000}{e^{-In(2)\times 20/10}}=40000s^{-1}$

Now,

$R_{o} = \lambda N_{o}$

$N_{o}=\frac{40000\: s^{-1}}{1.925\times 10^{-5}\; s^{-1}}$

$N_{o}= 2.08\times 10^{9}$

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