Question

The ratio of activities of two ratio nuclides $X$ and $Y$ in a mixture at time $t=0$ was found to be $4:1$. After two hours, the ratio of activities become $1:1$. If the $t_{1/2}$ of radio nuclide $X$ is $20$min then $t_{1/2}$ [in minutes] of radio nuclide $Y$ is :

• A. 10
• B. 20
• C. 30
• D. 40

Answer 1

The correct option is C 30

The activity of substance reduces to half of the initial value in one half life.

At $t=0$

Let activity of $X$ be ${4N}_0$

$\Rightarrow$ activity of $Y$ $=N_0$

$t_{1/2}$ of $X=20$ min

No. of half lives in 2 hours $=6$

Let the number of half lives of $Y$ in 2 hours be $x$.

At $t=$ 2hours

Activity of $X=\frac{{4N}_0}{{\left(2\right)}^6}$

Activity of $Y=\frac{N_0}{{\left(2\right)}^x}$

ATQ,

$\frac{4N_0}{2^6}=\frac{N_0}{2^x}$

$\Rightarrow x=4$

$\Rightarrow$ 4 half lives of $Y$ in 2 hours.

$\Rightarrow t_{1/2}forY=30min$