The half-life of a radioactive sample is T. If the activities of the sample at time t1 and t2 ( t1 < t2) are R1 and R2 respectively, then the number of atoms disintegrated in time t2 - t1 is proportional to

Activity R=λN so that R1=λN1and⁡R2=λN2R1−R2=λN1−N2=0.6931TN1−N2N1−N2=R−R2T0.6931N1−N2∝R1−R2T; ∴t2−t1αR1−R2T

The half-life of a radioactive sample is T. If the activities of the sample at time t1 and t2 ( t1 < t2) are R1 and R2 respectively, then the number of atoms disintegrated in time t2 - t1 is proportional to

Activity R=λN so that R1=λN1and⁡R2=λN2R1−R2=λN1−N2=0.6931TN1−N2N1−N2=R−R2T0.6931N1−N2∝R1−R2T; ∴t2−t1αR1−R2T

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The half-life of a radioactive sample is T. If the activities of the sample at time t₁ and t₂ ( t₁ < t₂) are R₁ and R₂ respectively, then the number of atoms disintegrated in time t₂ - t₁ is proportional to

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$(R_{1} + R_{2}) T$

$\frac{R_{1} + R_{2}}{T}$

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

$\frac{R_{1} R_{2}}{R_{1} + R_{2}} T$

answer is A.

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Detailed Solution

$Activity R = λN so that$
$\begin{array}{l} R_{1} = {λN}_{1} and R_{2} = {λN}_{2} \\ R_{1} - R_{2} = λ (N_{1} - N_{2}) = \frac{0 .6931}{T} (N_{1} - N_{2}) \\ N_{1} - N_{2} = \frac{(R - R_{2}) T}{0 .6931} \end{array}$
$N_{1} - N_{2} \propto (R_{1} - R_{2}) T; ∴ t_{2} - t_{1} α (R_{1} - R_{2}) T$