Question 60 of 311
An electron is released from rest near an infinite non-conducting sheet of uniform charge density ' '. The rate of change of de-Broglie wave length associated with the electron varies inversely as power of time. The numerical value of is ________.
Show answer and solution
Answer: 2
Let the momentum of at any time t is p and its de-broglie wavelength is .
Then,
$\begin{aligned}
& \frac{\mathrm{dp}}{\mathrm{dt}}=\frac{-\mathrm{h}}{\lambda^2} \frac{\mathrm{~d} \lambda}{\mathrm{dt}} \\
& \mathrm{ma}=\mathrm{F}=-\frac{\mathrm{h}}{\lambda} \frac{\mathrm{~d} \lambda}{\mathrm{dt}} \quad[\mathrm{~m}=\text { mass of } \mathrm{e}]
\end{aligned}$
Where, -ve sign represents decrease in with time
$\begin{aligned}
& \mathrm{a}=-\frac{\mathrm{p}^2}{\mathrm{mh}} \frac{\mathrm{~d} \lambda}{\mathrm{dt}} \\
& \mathrm{a}=-\frac{\mathrm{mv}^2}{\mathrm{~h}} \frac{\mathrm{~d} \lambda}{\mathrm{dt}} \\
& \frac{\mathrm{~d} \lambda}{\mathrm{dt}}=-\frac{\mathrm{ah}}{\mathrm{mv}^2}...(1)
\end{aligned}$
here,
and
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,
$\begin{aligned}
& \frac{\mathrm{dp}}{\mathrm{dt}}=\frac{-\mathrm{h}}{\lambda^2} \frac{\mathrm{~d} \lambda}{\mathrm{dt}} \\
& \mathrm{ma}=\mathrm{F}=-\frac{\mathrm{h}}{\lambda} \frac{\mathrm{~d} \lambda}{\mathrm{dt}} \quad[\mathrm{~m}=\text { mass of } \mathrm{e}]
\end{aligned}$
Where, -ve sign represents decrease in with time
$\begin{aligned}
& \mathrm{a}=-\frac{\mathrm{p}^2}{\mathrm{mh}} \frac{\mathrm{~d} \lambda}{\mathrm{dt}} \\
& \mathrm{a}=-\frac{\mathrm{mv}^2}{\mathrm{~h}} \frac{\mathrm{~d} \lambda}{\mathrm{dt}} \\
& \frac{\mathrm{~d} \lambda}{\mathrm{dt}}=-\frac{\mathrm{ah}}{\mathrm{mv}^2}...(1)
\end{aligned}$
here,
and
Substituting values of a \& v in equation (1)
$\begin{aligned}
& \frac{\mathrm{d} \lambda}{\mathrm{dt}}=-\frac{2 \mathrm{~h} \varepsilon_0}{\sigma \mathrm{et}^2} \\
& \Rightarrow \frac{\mathrm{~d} \lambda}{\mathrm{dt}} \propto \frac{1}{\mathrm{t}^2} \\
& \Rightarrow \mathrm{n}=2
\end{aligned}$