JEE Physics 2025 Question with Solution

Given: The electron has initial velocity $\vec{v}=v_0\hat{i}$ and enters the electric field $\vec{E}=-E_0\hat{k}$. The initial de Broglie wavelength is $\lambda_0=\frac{h}{mv_0}$.

Find: The de Broglie wavelength after time $t$.

Use the de Broglie relation:

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

Since the electric field is along $-\hat{k}$, the electron acquires a velocity component along $\hat{k}$ while its initial $\hat{i}$ component remains $v_0$ in magnitude.

The acceleration magnitude is:

$$a = \frac{eE_0}{m}$$

Hence the velocity after time $t$ is:

$$\vec{v}(t)=v_0\hat{i}-\frac{eE_0 t}{m}\hat{k}$$

Therefore, the speed is:

$$v(t)=\sqrt{v_0^2+\left(\frac{eE_0 t}{m}\right)^2}$$

So the new de Broglie wavelength becomes:

$$\lambda = \frac{h}{m\sqrt{v_0^2+\left(\frac{eE_0 t}{m}\right)^2}}$$

Using $\lambda_0=\frac{h}{mv_0}$,

$$\lambda = \lambda_0 \cdot \frac{v_0}{\sqrt{v_0^2+\left(\frac{eE_0 t}{m}\right)^2}}$$

Dividing numerator and denominator by $v_0$,

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

Therefore, the correct option is A.

Given: Initial momentum is along $\hat{i}$ and the electric field acts along $-\hat{k}$.

Find: The new de Broglie wavelength from the new momentum magnitude.

Initially,

$$p_0 = mv_0$$

and

$$\lambda_0 = \frac{h}{p_0}$$

The force due to the field changes the momentum in the $\hat{k}$ direction. After time $t$, the additional momentum magnitude is:

$$p_k = eE_0 t$$

Hence the resultant momentum magnitude is:

$$p = \sqrt{p_0^2+p_k^2}$$

Substituting,

$$p = \sqrt{m^2v_0^2+e^2E_0^2 t^2}$$

Now,

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

Using $h=\lambda_0 mv_0$,

$$\lambda = \frac{\lambda_0 mv_0}{\sqrt{m^2v_0^2+e^2E_0^2 t^2}}$$

Factor out $m^2v_0^2$ from the denominator:

$$\lambda = \lambda_0 \frac{mv_0}{mv_0\sqrt{1+\frac{e^2E_0^2 t^2}{m^2v_0^2}}}$$

Thus,

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

So the de Broglie wavelength decreases with time, and the correct option is A.