JEE Physics 2025 Question with Solution

Given: An electron is released from rest near an infinite non-conducting sheet of uniform charge density $-\sigma$.

Find: The numerical value of $n$ if the rate of change of de-Broglie wavelength varies inversely as the $n^{th}$ power of time.

For an infinite non-conducting sheet, the electric field has constant magnitude

$$E = \frac{\sigma}{2\epsilon_0}$$

So the acceleration of the electron is constant and is given by

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

If the electron starts from rest, then

$$v = at$$

Now de-Broglie wavelength is

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

where momentum $p = mv$. Hence

$$\lambda = \frac{h}{mv} = \frac{h}{m(at)}$$

Substituting the value of $a$,

$$\lambda = \frac{2\epsilon_0 h}{e\sigma t}$$

Differentiating with respect to time,

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

Therefore,

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

So the rate of change varies inversely as the second power of time.

Therefore, the numerical value of $n$ is $2$.

Given: The electron moves under constant acceleration due to the electric field of the sheet.

Find: How $\frac{d\lambda}{dt}$ depends on time.

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

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

Differentiating,

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

But

$$\frac{dp}{dt} = ma$$

So,

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

Using

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

we get

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

Since $p = mv$,

$$ma = -\frac{m^2 v^2}{h} \frac{d\lambda}{dt}$$

Therefore,

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

For the charged sheet,

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

And because the electron starts from rest,

$$v = at$$

Substituting $v = at$ in the expression for $\frac{d\lambda}{dt}$,

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

Hence,

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

So, the required value is $n = 2$.