The radioactivity of a sample is $R_{1}$ at a time $T_{1}$ and $R_{2}$ at a time $T_{2}$ . If the half-life of the specimen is T, the number of atoms that have disintegrated in the time $(T_{2} - T_{1})$ is equal to $\frac{n (R_{1} - R_{2}) T}{\ln 4}$ . Here n is some integral number. What is the value of n?

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answer is 2.

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Detailed Solution

$R_{1} = λ N_{1}, R_{2} = λ N_{2},$
No of atoms decayed in $(T_{1} - T_{2})$
$= N_{1} - N_{2} = \frac{R_{1} - R_{2}}{λ} = \frac{(R_{1} - R_{2}) T}{\ln 2} = \frac{2 (R_{1} - R_{2}) T}{\ln 4}$
Hence n=2

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