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Question

The activity of a sample is $A_{1}$ at a time $t_{1}$ and $A_{2}$ at time $t_{2}$ ( $t_{2} > t_{1}$ ). The half-life of sample is $T$ . Then, the number of atoms that have disintegrated in time ( $t_{1} - t_{2}$ ) is proportional to

\frac{A_{1}}{A_{2}}

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(A_{1} + A_{2})

T

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(A_{1} - A_{2})

T

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(\frac{A_{1} - A_{2}}{T})

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Solution

The correct option is D $(A_{1} - A_{2})$ $T$
If $N$ , is the total number of particles in the sample and $λ = l o g_{e} 2 / T$ is the decay constant.The activity— $A$ , is the number of decays per unit time of a radioactive sample.
It is given as $A = λ N \Rightarrow N = A / λ$
So at $t = t_{1}$ , $N_{1} = A_{1} / λ$
and at $t = t_{2}$ , $N_{2} = A_{2} / λ$
So number of disintegration is
$N_{1} - N_{2} = (A_{1} - A_{2}) / λ = (A_{1} - A_{2}) T / l o g_{e} 2 \propto (A_{1} - A_{2}) T$

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The activity of a sample is A1 at a time t1 and A2 at time t2(t2>t1). The half-life of sample is T. Then, the number of atoms that have disintegrated in time (t1−t2) is proportional to

The activity of a sample is $A_{1}$ at a time $t_{1}$ and $A_{2}$ at time $t_{2}$ ( $t_{2} > t_{1}$ ). The half-life of sample is $T$ . Then, the number of atoms that have disintegrated in time ( $t_{1} - t_{2}$ ) is proportional to