Question

A mixture consists of two radioactive materials $40g$ of $A_1$ and $160g$ of $A_2$ with half lives of $20s$ and $10s$ , respectively. Initially the mixture has $40gof{A}_1$ and $160gof{A}_2$ • The amount of the two in the mixture will become equal after

• A. $60s$
• B. $80s$
• C. $20s$
• D. $40s$

Answer 1

Answer: (D)

For $A_{1}and{A}_2$ ,
Initially ${\left(N_{A_1}\right)}_0=40g{\left(N_{A_2}\right)}_0=160g$
${\left(t_{1/2}\right)}_{A_1}=20s{\left(t_{1/2}\right)}_{A_2}=10s$
Suppos after time $T$ ,
$\left(N_{A_1}\right)=\left(N_{A_2}\right)\Rightarrow \frac{{\left(N_{A_1}\right)}_0}{2^{n_1}}=\frac{{\left(N_{A_2}\right)}_0}{2^{n_2}}\Rightarrow \frac{40}{160}=\frac{2^{n_1}}{2^{n_2}}\Rightarrow 2^{n_1-n_2}=2^{-2}{n}_2-n_1=2\left\{n=\frac{T}{t_{1/2}}\right\}\Rightarrow \frac{T}{10}-\frac{T}{20}=2\Rightarrow T=40s$