JEE Main 2023 — Chemistry Physical Chemistry

2023integereasy

A and B are two substances undergoing radioactive decay in a container. The half life of A is $15\mathrm{min}$ and that of B is $5\mathrm{min}$ . If the initial concentration of B is $4$ times that of A and they both start decaying at the same time, how much time will it take for the concentration of both of them to be same? ________min.

Official previous-year question

Held on 1 Feb 2023 · Verified 6 Jul 2026.

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Solution

For a first order reaction:

$\Rightarrow [A{]}_{t}=[A{]}_{0}{e}^{-\mathrm{kt}}$

For $A$ : Let $[A{]}_{t}$ be $y$ and $[A{]}_{0}$ be x

For a first order reaction:

$\therefore k=\frac{\mathrm{ln}2}{{t}_{1/2}}=\frac{\mathrm{ln}2}{15\mathrm{min}}$

$\Rightarrow y={\mathrm{xe}}^{-\mathrm{kt}}$

$={\mathrm{xe}}^{-(\frac{\mathrm{ln}2}{15})t}$ ....(1)

Similarly For B:

$[B{]}_{t}=[B{]}_{0}{e}^{-\mathrm{kt}}$

Let $[B{]}_{t}=y;[B{]}_{0}=4x; k=\frac{\mathrm{ln}2}{{t}_{1/2}}=\frac{\mathrm{ln}2}{5\mathrm{min}}$

$\Rightarrow y=4{\mathrm{xe}}^{-(\frac{\mathrm{ln}2}{5})t}$ ....(2)

Equating equation(1) and (2) we get:

$\Rightarrow {\mathrm{xe}}^{-(\frac{\mathrm{ln}2}{15})t}=4{\mathrm{xe}}^{-(\frac{\mathrm{ln}2}{5})t}$

$\Rightarrow t\times [\frac{\mathrm{ln}2}{5}-\frac{\mathrm{ln}2}{15}]=\mathrm{ln}4$

$\Rightarrow t\times \mathrm{ln}2[\frac{1}{5}-\frac{1}{15}]=2\mathrm{ln}2$

$\Rightarrow {e}^{t(\frac{\mathrm{ln}2}{5}-\frac{\mathrm{ln}2}{15})}=4$

$\Rightarrow t=15\mathrm{min}$

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Solution

For a first order reaction:

$\Rightarrow [A{]}_{t}=[A{]}_{0}{e}^{-\mathrm{kt}}$

For $A$ : Let $[A{]}_{t}$ be $y$ and $[A{]}_{0}$ be x

For a first order reaction:

$\therefore k=\frac{\mathrm{ln}2}{{t}_{1/2}}=\frac{\mathrm{ln}2}{15\mathrm{min}}$

$\Rightarrow y={\mathrm{xe}}^{-\mathrm{kt}}$

$={\mathrm{xe}}^{-(\frac{\mathrm{ln}2}{15})t}$ ....(1)

Similarly For B:

$[B{]}_{t}=[B{]}_{0}{e}^{-\mathrm{kt}}$

Let $[B{]}_{t}=y;[B{]}_{0}=4x; k=\frac{\mathrm{ln}2}{{t}_{1/2}}=\frac{\mathrm{ln}2}{5\mathrm{min}}$

$\Rightarrow y=4{\mathrm{xe}}^{-(\frac{\mathrm{ln}2}{5})t}$ ....(2)

Equating equation(1) and (2) we get:

$\Rightarrow {\mathrm{xe}}^{-(\frac{\mathrm{ln}2}{15})t}=4{\mathrm{xe}}^{-(\frac{\mathrm{ln}2}{5})t}$

$\Rightarrow t\times [\frac{\mathrm{ln}2}{5}-\frac{\mathrm{ln}2}{15}]=\mathrm{ln}4$

$\Rightarrow t\times \mathrm{ln}2[\frac{1}{5}-\frac{1}{15}]=2\mathrm{ln}2$

$\Rightarrow {e}^{t(\frac{\mathrm{ln}2}{5}-\frac{\mathrm{ln}2}{15})}=4$

$\Rightarrow t=15\mathrm{min}$

JEE Main 2023 — Chemistry Physical Chemistry

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Solution

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JEE Main 2023 — Chemistry Physical Chemistry

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry