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Rutherford's Set up and Predictions

The activity ...

Question

The activity of a radioactive sample is $A_{1}$ at time $t_{1}$ and $A_{2}$ at time $t_{2}$ . If $T_{a v}$ is the average life time of the radioactive sample, then the number of atoms that decay during interval $(t_{2} - t_{1})$ is

A_{1} t_{1} - A_{2} t_{2}

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(A_{1} - A_{2}) T_{a v}

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\frac{A_{1} - A_{2}}{T_{a v}}

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

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Solution

The correct option is B $(A_{1} - A_{2}) T_{a v}$
$A c t i v i t y, A = λ N \Rightarrow A_{1} = λ N_{1} = \frac{1}{T_{a v}} N_{1}; A_{2} = \frac{N_{2}}{T_{a v}} ∴ (N_{1} - N_{2}) = (A_{1} - A_{2}) T_{a v}$

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The activity of a radioactive sample is A1 at time t1 and A2 at time t2. If Tav is the average life time of the radioactive sample, then the number of atoms that decay during interval (t2−t1) is

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The activity of a radioactive sample is $A_{1}$ at time $t_{1}$ and $A_{2}$ at time $t_{2}$ . If $T_{a v}$ is the average life time of the radioactive sample, then the number of atoms that decay during interval $(t_{2} - t_{1})$ is

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