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Two radioactive sources $A$ and $B$ initially contain equal number of radioactive atoms. Source $A$ has a half-life of 2 hour and source $B$ has a half-life of 4 hours. At the end of 4 hours, the ratio of the rate of disintegration of $B$ to that of $A$ is :
Option: 1
$1: 2$

Option: 2
$2: 1$

Option: 3
$1: 1$

Option: 4
$1: 4$

Answers (1)

best_answer

Two radioactive nuclei $\mathrm{A}$ and $\mathrm{B}$ initially contain equal number of atoms. The half life is 2 hour and 4 hours respectively. Given, nucleii A and B contain equal number of atoms.

Half-life of $A,\left(\mathrm{~T}_{\frac{1}{2}}\right)_1=2 \mathrm{hr}$

Half-life of $\mathrm{Y},\left(\mathrm{T}_{\frac{1}{2}}\right)_2=4 \mathrm{hr}$

Radio of radioactive sample of A left after 4 hours,

$\mathrm{N}_1=\left(\frac{1}{2}\right)^{4 / 2} \mathrm{~N}_0=\frac{1}{4} \mathrm{~N}_0$

Radio of radioactive sample of B left after 4 hours,

$\mathrm{N}_2=\left(\frac{1}{2}\right)^{4 / 4} \mathrm{~N}_0=\frac{1}{2} \mathrm{~N}_0$

Now, decay rate is given by,

$\mathrm{R}=\lambda \mathrm{N}=\frac{0.693 \mathrm{~N}}{\mathrm{~T}_{12}}$

$\Rightarrow \mathrm{R} \propto \frac{\mathrm{N}}{\mathrm{T}_{1 / 2}}$

$\frac{R_2}{R_1}=\frac{\left(\mathrm{T}_{\frac{1}{2}}\right)_1}{\left(\mathrm{~T}_{\frac{1}{2}}\right)_2} \times \frac{\mathrm{N}_2}{\mathrm{~N}_1}$

$\text { i.e., }=\frac{R_2}{R_1}=\frac{2}{4} \times \frac{\frac{\mathrm{N}_0}{2}}{\frac{\mathrm{N}_0}{4}}$

$=1$

Posted by

Gaurav

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