JEE Mains · Physics · STD 12 - Dual nature of radiation and matter
- JEE Mains 2025
- Hard
- Numerical
An electron is released from rest near an infinite non-conducting sheet of uniform charge density ' \(-\sigma\) '. The rate of change of de-Broglie wave length associated with the electron varies inversely as \(\mathrm{n}^{\text {th }}\) power of time. The numerical value of \(n\) is ________.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) 2
Step-by-step Solution
Detailed explanation
Let the momentum of \(\mathrm{e}^{-}\)at any time t is p and its de-broglie wavelength is \(\lambda\).
Then, \(\mathrm{p}=\frac{\mathrm{h}}{\lambda}\)
\(\begin{aligned}
& \frac{\mathrm{dp}}{\mathrm{dt}}=\frac{-\mathrm{h}}{\lambda^2} \frac{\mathrm{~d} \lambda}{\mathrm{dt}} \\
& \mathrm{ma}=\mathrm{F}=-\frac{\mathrm{h}}{\lambda} \frac{\mathrm{~d} \lambda}{\mathrm{dt}} \quad[\mathrm{~m}=\text { mass of } \mathrm{e}]
\end{aligned}\)
Where, -ve sign represents decrease in \(\lambda\) with time
\(\mathrm{ma}=\frac{-\mathrm{h}}{(\mathrm{~h} / \mathrm{p})^2} \frac{\mathrm{~d} \lambda}{\mathrm{dt}}\)
\(\begin{aligned}
& \mathrm{a}=-\frac{\mathrm{p}^2}{\mathrm{mh}} \frac{\mathrm{~d} \lambda}{\mathrm{dt}} \\
& \mathrm{a}=-\frac{\mathrm{mv}^2}{\mathrm{~h}} \frac{\mathrm{~d} \lambda}{\mathrm{dt}} \\
& \frac{\mathrm{~d} \lambda}{\mathrm{dt}}=-\frac{\mathrm{ah}}{\mathrm{mv}^2}...(1)
\end{aligned}\)
here, \(\mathrm{a}=\frac{\mathrm{qE}}{\mathrm{m}}=\frac{\mathrm{e}}{\mathrm{m}} \frac{\sigma}{2 \varepsilon_0}\)
\(\mathrm{a}=\frac{\sigma \mathrm{e}}{2 \mathrm{~m} \varepsilon_0}\)
and \(\mathrm{v}=\mathrm{u}+\mathrm{at}\)
\(\mathrm{v}=\mathrm{at}\)
Substituting values of a \& v in equation (1)
\(\begin{aligned}
& \frac{\mathrm{d} \lambda}{\mathrm{dt}}=-\frac{2 \mathrm{~h} \varepsilon_0}{\sigma \mathrm{et}^2} \\
& \Rightarrow \frac{\mathrm{~d} \lambda}{\mathrm{dt}} \propto \frac{1}{\mathrm{t}^2} \\
& \Rightarrow \mathrm{n}=2
\end{aligned}\)
Then, \(\mathrm{p}=\frac{\mathrm{h}}{\lambda}\)
\(\begin{aligned}
& \frac{\mathrm{dp}}{\mathrm{dt}}=\frac{-\mathrm{h}}{\lambda^2} \frac{\mathrm{~d} \lambda}{\mathrm{dt}} \\
& \mathrm{ma}=\mathrm{F}=-\frac{\mathrm{h}}{\lambda} \frac{\mathrm{~d} \lambda}{\mathrm{dt}} \quad[\mathrm{~m}=\text { mass of } \mathrm{e}]
\end{aligned}\)
Where, -ve sign represents decrease in \(\lambda\) with time
\(\mathrm{ma}=\frac{-\mathrm{h}}{(\mathrm{~h} / \mathrm{p})^2} \frac{\mathrm{~d} \lambda}{\mathrm{dt}}\)
\(\begin{aligned}
& \mathrm{a}=-\frac{\mathrm{p}^2}{\mathrm{mh}} \frac{\mathrm{~d} \lambda}{\mathrm{dt}} \\
& \mathrm{a}=-\frac{\mathrm{mv}^2}{\mathrm{~h}} \frac{\mathrm{~d} \lambda}{\mathrm{dt}} \\
& \frac{\mathrm{~d} \lambda}{\mathrm{dt}}=-\frac{\mathrm{ah}}{\mathrm{mv}^2}...(1)
\end{aligned}\)
here, \(\mathrm{a}=\frac{\mathrm{qE}}{\mathrm{m}}=\frac{\mathrm{e}}{\mathrm{m}} \frac{\sigma}{2 \varepsilon_0}\)
\(\mathrm{a}=\frac{\sigma \mathrm{e}}{2 \mathrm{~m} \varepsilon_0}\)
and \(\mathrm{v}=\mathrm{u}+\mathrm{at}\)
\(\mathrm{v}=\mathrm{at}\)
Substituting values of a \& v in equation (1)
\(\begin{aligned}
& \frac{\mathrm{d} \lambda}{\mathrm{dt}}=-\frac{2 \mathrm{~h} \varepsilon_0}{\sigma \mathrm{et}^2} \\
& \Rightarrow \frac{\mathrm{~d} \lambda}{\mathrm{dt}} \propto \frac{1}{\mathrm{t}^2} \\
& \Rightarrow \mathrm{n}=2
\end{aligned}\)
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