Two radioactive samples A and B have half lives T₁ and T₂left(T₁ > T₂right) respectively

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13.Nuclei

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Two radioactive samples $A$ and $B$ have half lives $T_1$ and $T_2\left(T_1 > T_2\right)$ respectively At $t=0$, the activity of $B$ was twice the activity of $A$. Their activity will become equal after a time

A

$\frac{T_1 T_2}{T_1-T_2}$

B

$\frac{T_1-T_2}{2}$

C

$\frac{T_1+T_2}{2}$

D

$\frac{T_1 T_2}{T_1+T_2}$

Solution

(a)

$2 R_A=R_B$

$2 \lambda_1 N_1=\lambda_2 N_2 \quad \dots (i)$

Radio-activity is same after say time $t$

$\lambda_1 N_1 e^{-\lambda_1 t}=\lambda_2 N_2 e^{-\lambda_2 t} \quad \dots (ii)$

Dividing $(i)$ by $(ii)$

$2 e^{\lambda_1 t}=e^{\lambda_2 t}$

$2=e^{\left(\lambda_2-\lambda_1\right) t}$

Taking In on both sides

$0.693=\left(\lambda_2-\lambda_1\right) t$

$1=\left(\frac{1}{T_2}-\frac{1}{T_1}\right) t$

$\frac{T_2 T_1}{T_1-T_2}=t$

Standard 12

Physics

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