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

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, \( n = 2 \)

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