Radium disappears at a rate proportional to the amount present. If $5 %$ of the original amount disappears in $50$ years, how much will remain at the end of 100 years. [Take $A_{0}$ as the initial amount].

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Solution

Let "x" be the quantity of radium at time "t".

Given, $\frac{d x}{d t} \propto x$

$\Rightarrow \frac{d x}{d t} = k x$

$\frac{d x}{x} = k d t$

Integrating on both sides, we get,

$log x = k t + log c$

$log \frac{x}{c} = k t$

$∴ x = c e^{k t}$

When $t = 0, x = A_{0}$

$A_{0} = c e^{k \times 0}$

$∴ c = A_{0}$

When $t = 50, x = 0.95 A_{0}$

$0.95 A_{0} = A_{0} e^{50 k}$

$∴ e^{50 k} = 0.95$

When $t = 100$

$x = A_{0} e^{100 k}$

$= A_{0} e^{50 k} e^{50 k}$

$= A_{0} (0.95) (0.95)$

$= 0.9025 A_{0}$

The amount of radium remaining after 100 years is $= 0.9025 A_{0}$

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