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:

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