The radioactivity of sample is $R_{1}$ at a time $T_{1}$ and Iii 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 :

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$(R_{1} - R_{2}) T / 0.693$

$(R_{1} - R_{2})$

$(R_{1} T_{1} - R_{2} T_{2})$

$(R_{1} - R_{2}) / T$

answer is D.

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Detailed Solution

$Radioactivity = \frac{dN}{dt} = λ N$

$\begin{aligned} R_{1} = λ N (T_{1}) \\ R_{2} = λ N (T_{2}) \end{aligned}$

Number of fragmented atoms

$= N (T_{2}) - N (T_{1}) = \frac{R_{2} - R_{1}}{λ} = \frac{(R_{2} - R_{1}) T}{\ln 2}$

Given

$R_{1} = k N_{1}; R_{2} = k N_{2}$

$\Rightarrow N_{1} - N_{2} = \frac{R_{1} - R_{2}}{K} = \frac{T}{0.693} [R_{1} - R_{2}]$

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