The activity ...

Question

The activity of sample of radioactive materials is $A_{1}$ at time $t_{1}$ and $A_{2}$ at time $t_{2} (t_{2} > t_{1})$ . It's mean life is $T$ , then

A_{2} = A_{1} e^{⎛ ⎝ \frac{t_{1} - t_{2}}{T} ⎞ ⎠}

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A_{2} = A_{1} e^{⎛ ⎜ ⎝ \frac{t_{1} / t_{2}}{T} ⎞ ⎟ ⎠}

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

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\frac{A_{1} - A_{2}}{t_{2} - t_{1}} = Constant

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Solution

The correct option is A $A_{2} = A_{1} e^{⎛ ⎝ \frac{t_{1} - t_{2}}{T} ⎞ ⎠}$
According to the radioactive decay law,

$A_{1} = A_{0} e^{- λ t_{1}} . . . (1)$

$A_{2} = A_{0} e^{- λ t_{2}} . . . (2)$

Using equations $(1)$ and $(2)$ ,

$\frac{A_{1}}{A_{2}} = \frac{A_{0} e^{- λ t_{1}}}{A_{0} e^{- λ t_{2}}} = e^{- λ (t_{1} - t_{2})}$

$\Rightarrow \frac{A_{1}}{A_{2}} = e^{- \frac{(t_{1} - t_{2})}{T}} [∵ T = \frac{1}{λ}]$

$\Rightarrow \frac{A_{2}}{A_{1}} = e^{⎛ ⎝ \frac{t_{1} - t_{2}}{T} ⎞ ⎠}$

$∴ A_{2} = A_{1} e^{⎛ ⎝ \frac{t_{1} - t_{2}}{T} ⎞ ⎠}$

Hence, $(A)$ is the correct answer.

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