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 ______.

Hint:

- 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.

Answer 1

Correct Answer: 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: \[ \left|\frac{d\lambda}{dt}\right| \propto \frac{1}{t^2} \quad \Rightarrow \quad n = 2 \]