Question

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

• A. $20\mathrm{s}$
• B. $60\mathrm{s}$
• C. $80\mathrm{s}$
• D. $40\mathrm{s}$

Answer 1

The correct option is D

$40\mathrm{s}$

Explanation for correct option:

(D) $40\mathrm{s}$

Step 1: Formula:

$$\begin{gathered} \mathrm{Let}\mathrm{after}\mathrm{t}\mathrm{s}\mathrm{amount}\mathrm{of}{\mathrm{A}}_1\mathrm{and}{\mathrm{A}}_2\mathrm{will}\mathrm{become}\mathrm{equal}\mathrm{in}\mathrm{mixture} \\ \mathrm{N}={\mathrm{N}}_0{\left(\frac{1}{2}\right)}^{\mathrm{n}} \\ \mathrm{Where},\mathrm{n}\mathrm{is}\mathrm{the}\mathrm{half}\mathrm{life}. \\ \mathrm{For}{\mathrm{A}}_1,{\mathrm{N}}_1={\mathrm{N}}_{01}{\left(\frac{1}{2}\right)}^{\raisebox{1ex}{$\mathrm{t}$}\!\left/ \!\raisebox{-1ex}{$20$}\right.} \\ \mathrm{For}{\mathrm{A}}_2,{\mathrm{N}}_2={\mathrm{N}}_{02}{\left(\frac{1}{2}\right)}^{\raisebox{1ex}{$\mathrm{t}$}\!\left/ \!\raisebox{-1ex}{$10$}\right.} \end{gathered}$$

Step 2: Equalising the amount:

$$\begin{gathered} \mathrm{According}\mathrm{to}\mathrm{the}\mathrm{question}, \\ {\mathrm{N}}_1={\mathrm{N}}_2 \\ \frac{40}{2^{{\displaystyle \raisebox{1ex}{$\mathrm{t}$}\!\left/ \!\raisebox{-1ex}{$20$}\right.}}}=\frac{160}{2^{{\displaystyle \raisebox{1ex}{$\mathrm{t}$}\!\left/ \!\raisebox{-1ex}{$10$}\right.}}} \\ \frac{1}{2^{{\displaystyle \raisebox{1ex}{$\mathrm{t}$}\!\left/ \!\raisebox{-1ex}{$20$}\right.}}}=\frac{4}{2^{{\displaystyle \raisebox{1ex}{$\mathrm{t}$}\!\left/ \!\raisebox{-1ex}{$10$}\right.}}} \\ {2}^{\raisebox{1ex}{$\mathrm{t}$}\!\left/ \!\raisebox{-1ex}{$10$}\right.}=4\mathrm{x}{2}^{\raisebox{1ex}{$\mathrm{t}$}\!\left/ \!\raisebox{-1ex}{$20$}\right.} \\ {2}^{\raisebox{1ex}{$\mathrm{t}$}\!\left/ \!\raisebox{-1ex}{$10$}\right.}=2^2\mathrm{x}{2}^{\raisebox{1ex}{$\mathrm{t}$}\!\left/ \!\raisebox{-1ex}{$20$}\right.} \end{gathered}$$

Step 3: Finding the time required to achieve equal amount:

$$\begin{gathered} 2^{\raisebox{1ex}{$\mathrm{t}$}\!\left/ \!\raisebox{-1ex}{$10$}\right.}=2^2.2^{\raisebox{1ex}{$\mathrm{t}$}\!\left/ \!\raisebox{-1ex}{$20$}\right.} \\ \mathrm{Adding}\mathrm{powers}\mathrm{as}\mathrm{the}\mathrm{base}\mathrm{is}\mathrm{equal}, \\ {2}^{\raisebox{1ex}{$\mathrm{t}$}\!\left/ \!\raisebox{-1ex}{$10$}\right.}=2^{\left({}^{(\raisebox{1ex}{$\mathrm{t}$}\!\left/ \!\raisebox{-1ex}{$20$}\right.+2}\right)} \\ \frac{\mathrm{t}}{10}=\frac{\mathrm{t}}{20}+2 \\ \frac{\mathrm{t}}{10}-\frac{\mathrm{t}}{20}=2 \\ \frac{2\mathrm{t}-\mathrm{t}}{20}=2 \\ \frac{\mathrm{t}}{20}=2 \\ \mathrm{t}=40\mathrm{s} \end{gathered}$$

So, this is the correct option.

Explanation for incorrect options:

Since, the correct answer is $40\mathrm{s}$. Therefore, the options (A), (B) and (C) stand incorrect.

Hence, the correct option is (D) $40\mathrm{s}$