Question:

An electron is released from rest near an infinite non-conducting sheet of uniform charge density '–σ'. The rate of change of de-Broglie wavelength associated with the electron varies inversely as $n^{th}$ power of time. The numerical value of $n$ is ______.

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- For problems involving charged sheets, remember the electric field is constant: $E = \frac{\sigma}{2\epsilon_0}$. - The de-Broglie wavelength is inversely proportional to momentum: $\lambda = \frac{h}{p}$. - When dealing with rates, carefully differentiate and observe power-law relationships.

Updated On: Jan 31, 2026

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Correct Answer: 2

Approach Solution - 1

Let the momentum of the electron at any time $ t $ be $ p $, and its de-Broglie wavelength is $ \lambda $.

Then, the momentum is given by:

$ p = \frac{h}{\lambda} $

The rate of change of momentum is:

$ \frac{dp}{dt} = -\frac{h}{\lambda^2} \frac{d\lambda}{dt} $

Now, using $ ma = F $ (where $ m $ is the mass of the electron), we have:

$ ma = -\frac{h}{\lambda} \frac{d\lambda}{dt} $

Where the negative sign represents a decrease in $ \lambda $ with time.

From the above equation:

$ ma = -\frac{h}{\left( \frac{h}{p} \right)^2} \frac{d\lambda}{dt} $

Simplifying further:

$ a = -\frac{p^2}{mh} \frac{d\lambda}{dt} $

$ a = -\frac{mv^2}{h} \frac{d\lambda}{dt} $

From this, we can find:

$ \frac{d\lambda}{dt} = -\frac{ah}{mv^2} \quad \cdots (1) $

Here, $ a = \frac{qE}{m} = \frac{e}{m} \frac{\sigma}{2\epsilon_0} $

Also, $ v = u + at $, and since $ u = 0 $, we have:

$ v = at $

Substituting the values of $ a $ and $ v $ in equation (1):

$ \frac{d\lambda}{dt} = -\frac{2h\epsilon_0}{\sigma e t^2} $

Thus, we get:

$ \frac{d\lambda}{dt} \propto \frac{1}{t^2} $

Therefore, $ n = 2 $

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Approach Solution -2

Step 1: Determine the electric field and force on the electron.
- For an infinite non-conducting sheet with charge density $-\sigma$, the electric field $E$ is: \[ E = \frac{\sigma}{2\epsilon_0} \] - The force on the electron (charge $-e$) is: \[ F = -eE = -\frac{e\sigma}{2\epsilon_0} \] - The acceleration $a$ of the electron is: \[ a = \frac{F}{m_e} = -\frac{e\sigma}{2\epsilon_0 m_e} \]
Step 2: Find the velocity as a function of time. - Since the electron starts from rest, its velocity $v$ at time $t$ is: \[ v = at = -\frac{e\sigma}{2\epsilon_0 m_e} t \]
Step 3: Express the de-Broglie wavelength $\lambda$. - The de-Broglie wavelength is given by: \[ \lambda = \frac{h}{p} = \frac{h}{m_e v} = \frac{h}{m_e \left|\frac{e\sigma}{2\epsilon_0 m_e} t\right|} = \frac{2\epsilon_0 h}{e\sigma t} \]
Step 4: Compute the rate of change of $\lambda$ with respect to time. \[ \frac{d\lambda}{dt} = -\frac{2\epsilon_0 h}{e\sigma t^2} \] - The magnitude of the rate of change is: \[ \left|\frac{d\lambda}{dt}\right| \propto \frac{1}{t^2} \]
Step 5: Compare with the given relation. - The problem states that $\frac{d\lambda}{dt}$ varies inversely as the $n^{th}$ power of time.
From Step 4, we see: \[ \left|\frac{d\lambda}{dt}\right| \propto \frac{1}{t^2} \quad \Rightarrow \quad n = 2 \]

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