QJEEM25Ph0225: JEE Main 2025 Physics - Class 12 JEEM25Ph0225 Solution: An electron of mass ' m ' with an initial velocity enters an…

Derivation of the de Broglie wavelength after time $t$

The de Broglie wavelength is given by

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

where $h$ is Planck's constant and $p$ is the momentum of the particle.

Initial Conditions:

Initially, the electron has a velocity

$\vec{v} = v_0\,\hat{i},$

so its momentum is

$p_0 = m v_0.$

The initial de Broglie wavelength is therefore

$\lambda_0 = \frac{h}{m v_0}.$

Effect of the Electric Field:

The electron enters an electric field

$\vec{E} = -E_0\,\hat{k}.$

For an electron with charge $-e$ , the force is given by

$\vec{F} = -e\,\vec{E} = -e\left(-E_0\,\hat{k}\right) = e E_0\,\hat{k}.$

This force causes an acceleration in the $\hat{k}$ direction:

$a_z = \frac{F_z}{m} = \frac{e E_0}{m}.$

Momentum Components After Time $t$ :

In the $\hat{i}$ direction, there is no force, so the momentum remains constant:

$p_x = m v_0.$

In the $\hat{k}$ direction, starting from rest, the momentum becomes:

$p_z = m v_z = m \left(a_z t\right) = e E_0 t.$

Total Momentum of the Electron:

The total momentum is the vector sum of the components:

$p = \sqrt{p_x^2 + p_z^2} = \sqrt{(m v_0)^2 + (e E_0 t)^2}.$

New de Broglie Wavelength:

Substituting the expression for the total momentum into the de Broglie relation gives:

$\lambda = \frac{h}{\sqrt{m^2 v_0^2 + e^2 E_0^2 t^2}}.$

Expressing this in terms of the initial de Broglie wavelength $\lambda_0 = \frac{h}{m v_0}$ :

$\lambda = \frac{\lambda_0}{\sqrt{1+\frac{e^2 E_0^2 t^2}{m^2 v_0^2}}}.$

This final expression matches the option:

$\frac{\lambda_0}{\sqrt{1+\frac{e^2 E_0^2 t^2}{m^2 v_0^2}}}.$