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Standard XII

Chemistry

Relation between Cp and Cv

The half life...

Question

The half life of a ratio active source is T. If radio activities of the samples are $R_{1}$ and $R_{2}$ at time $T_{1}$ and $T_{2}$ respectively, then the number of atoms disintegrated in time $(T_{2} - T_{1})$ is proportional to

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

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

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\frac{R_{1} - R - 2}{T}

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\frac{R_{1} + R - 2}{T}

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Solution

The correct option is B $(R_{1} - R_{2}) T$
$R_{1} = N_{1} λ$
$R_{2} = N_{2} λ$
$\frac{R_{1} - R_{2}}{λ} \propto N_{1} - N_{2}$

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The half life of a ratio active source is T. If radio activities of the samples are R1 and R2 at time T1 and T2 respectively, then the number of atoms disintegrated in time (T2−T1) is proportional to

The correct option is B (R1−R2)TR1=N1λ R2=N2λ R1−R2λ∝N1−N2

The half life of a ratio active source is T. If radio activities of the samples are $R_{1}$ and $R_{2}$ at time $T_{1}$ and $T_{2}$ respectively, then the number of atoms disintegrated in time $(T_{2} - T_{1})$ is proportional to

The correct option is B $(R_{1} - R_{2}) T$
$R_{1} = N_{1} λ$
$R_{2} = N_{2} λ$
$\frac{R_{1} - R_{2}}{λ} \propto N_{1} - N_{2}$