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 $40gofA_{1}$ and $160gofA_{2}$ • 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 $A_{1}andA_{2}$ ,
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 $T$ ,
$\left(N_{A_{1}}\right)=\left(N_{A_{2}}\right)\Rightarrow \frac{\left(N_{A_{1}}\right)_{0}}{2^{n_{1}}}=\frac{\left(N_{A_{2}}\right)_{0}}{2^{n_{2}}}\Rightarrow \frac{40}{160}=\frac{2^{n_{1}}}{2^{n_{2}}}\Rightarrow 2^{n_{1} - n_{2}}=2^{- 2}n_{2}-n_{1}=2\left\{n = \frac{T}{t_{1/2}}\right\}\Rightarrow \frac{T}{10}-\frac{T}{20}=2\Rightarrow T=40s$