A radioactive isotope is being produced at a constant rate $X$ . Half-life of the radioactive substance is $Y$ . After some time, the number of radioactive nuclei become constant. The value of this constant is

\frac{X Y}{l n (2)}

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X Y

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(X Y) l n (2)

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\frac{X}{Y}

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Solution

The correct option is A $\frac{X Y}{l n (2)}$
At any given time $t$ :
$\frac{d N}{d t} = X - λ N$
where, $λ = \frac{l n 2}{Y}$
After time $t$ , $N =$ const.
$\frac{d N}{d t} = 0$
$X - λ N$ = 0
$N =$ $\frac{X Y}{ln 2}$

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