Two radioactive elements $X$ and $Y$ have half-lives of 50 and 100 minute respectively. An initial sample of both the elements has same no. of atoms. The ratio of the remaining number of atoms of $X$ and $Y$ after 200 minutes is:

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1/2

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1/4

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Solution

The correct option is D 1/4
As we know,
$N = N_{0} / 2^{n}$ no. of particles after n half lives
here,
200 minutes means 4 half life of X nad 2 half life of Y.
$A m o u n t l e f t o f X = \frac{N_{0}}{2^{4}} ∵ n = 4$ (half-lives)
$A m o u n t l e f t o f Y = \frac{N_{0}}{2^{2}} ∵ n = 2$
$∴ R a t i o o f X : Y l e f t = \frac{2^{2}}{2^{4}} = \frac{1}{4}$

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Two radioactive elements X and Y have half-lives of 50 and 100 minute respectively. An initial sample of both the elements has same no. of atoms. The ratio of the remaining number of atoms of X and Y after 200 minutes is:

The correct option is D 1/4As we know,N=N0/2n no. of particles after n half liveshere,200 minutes means 4 half life of X nad 2 half life of Y. AmountleftofX=N024∵n=4 (half-lives) AmountleftofY=N022∵n=2 ∴RatioofX:Yleft=2224=14

Two radioactive elements $X$ and $Y$ have half-lives of 50 and 100 minute respectively. An initial sample of both the elements has same no. of atoms. The ratio of the remaining number of atoms of $X$ and $Y$ after 200 minutes is: