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Alpha Decay

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Question

The activity of a sample of a radioactive material is $A_{1}$ at time $t_{1}$ and $A_{2}$ at time $t_{2} (t_{2} > t_{1})$ . If its mean is life T, then

A

A_{1} t_{1} = A_{2} t_{2}

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B

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

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C

A_{2} = A_{1} e^{\frac{(t_{1} - t_{2})}{T}}

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D

A_{2} = A_{1} e^{\frac{(t_{1} - t_{2})}{T}}

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Solution

The correct option is C $A_{2} = A_{1} e^{\frac{(t_{1} - t_{2})}{T}}$
We know that actvity of a radioactive material is written as
$A = A_{0} e^{\frac{- t}{τ}}$

So for the given conditions in question,
$A_{1} = A_{0} e^{\frac{- t_{1}}{T}}$
$A_{2} = A_{0} e^{\frac{- t_{2}}{T}}$

Now,
$\frac{A_{2}}{A_{1}} = \frac{A_{0} e^{\frac{- t_{2}}{T}}}{A_{0} e^{\frac{- t_{1}}{T}}}$
$∴ A_{2} = A_{1} e^{\frac{t_{1} - t_{2}}{T}}$

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