JEE Main 2026 — Dual Nature of Matter Question with Solution

The force experienced by the electron in the electric field is given by $\vec{F}=q\vec{E}$.

Since the charge of an electron is $-e$ and the electric field is $\vec{E}=-2E_0\hat{i}$, the force is:

$\vec{F}=(-e)(-2E_0\hat{i})=2e{E}_0\hat{i}$

The acceleration of the electron is:

$\vec{a}=\frac{\vec{F}}{m}=\frac{2e{E}_0}{m}\hat{i}$

Using the first equation of motion, the velocity of the electron at time $t$ is:

$\vec{v}(t)=\vec{V}+\vec{a}t=\left(v_0+\frac{2e{E}_0}{m}t\right)\hat{i}$

The de Broglie wavelength of the electron at time $t$ is:

$\lambda =\frac{h}{m|\vec{v}(t)|}=\frac{h}{m\left(v_0+\frac{2e{E}_0}{m}t\right)}$

We are given that ${\lambda }_0=\frac{h}{4m{v}_0}$, which implies $h=4m{v}_0{\lambda }_0$. Substituting this into the expression for $\lambda$:

$\lambda =\frac{4m{v}_0{\lambda }_0}{m\left(v_0+\frac{2e{E}_0}{m}t\right)}$

Dividing the numerator and the denominator by $m{v}_0$, we get:

$\lambda =\frac{4{\lambda }_0}{1+\frac{2E_{0}e}{m}\frac{t}{v_0}}$

Answer: $\frac{4{\lambda }_0}{\left[1+\frac{2E_{0}e}{m}\frac{t}{v_0}\right]}$