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Half-life of a radioactive substance is T. At time $t_{1}$ activity of a radioactive substance is $R_{1}$ and at time $t_{2}$ it is $R_{2}$ . Find the number of nuclei decayed in this interval of time.

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$N_{1} - N_{2} = \frac{(R_{1} - R_{2}) T}{\ln (2)}$

$N_{2} - N_{1} = \frac{(R_{1} - R_{2}) T}{\ln (2)}$

$N_{1} - N_{2} = \frac{(R_{2} - R_{1}) T}{\ln (2)}$

$N_{2} - N_{1} = \frac{(R_{2} - R_{1}) T}{\ln (2)}$

answer is A.

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Detailed Solution

Half-life is given by

$t_{\frac{1}{2}} = \frac{\ln (2)}{λ} λ = \frac{\ln (2)}{t_{\frac{1}{2}}} = \frac{\ln (2)}{T}$

Activity $R = λ N$

$N = \frac{R}{λ} = \frac{R T}{\ln (2)}$

When activity is R1, numbers of nuclei are

$N_{1} = \frac{R_{1} T}{\ln (2)}$

Similarly, $N_{2} = \frac{R_{2} T}{\ln (2)}$

So, Numbers decayed

$= N_{1} - N_{2} = \frac{(R_{1} - R_{2}) T}{\ln (2)}$

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