JEE Main 2026 — Physics Modern Physics

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

Since the charge of an electron is $-e$−e and the electric field is $\vec{E} = -2E_0\hat{i}$E=−2E0​i^, the force is:

$$\vec{F} = (-e)(-2E_0\hat{i}) = 2eE_0\hat{i}$$F=(−e)(−2E0​i^)=2eE0​i^

The acceleration of the electron is:

$$\vec{a} = \dfrac{\vec{F}}{m} = \dfrac{2eE_0}{m}\hat{i}$$a=mF​=m2eE0​​i^

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

$$\vec{v}(t) = \vec{V} + \vec{a}t = \left(v_0 + \dfrac{2eE_0}{m}t\right)\hat{i}$$v(t)=V+at=(v0​+m2eE0​​t)i^

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

$$\lambda = \dfrac{h}{m|\vec{v}(t)|} = \dfrac{h}{m\left(v_0 + \dfrac{2eE_0}{m}t\right)}$$λ=m∣v(t)∣h​=m(v0​+m2eE0​​t)h​

We are given that $\lambda_0 = \dfrac{h}{4mv_0}$λ0​=4mv0​h​, which implies $h = 4mv_0\lambda_0$h=4mv0​λ0​. Substituting this into the expression for $\lambda$λ:

$$\lambda = \dfrac{4mv_0\lambda_0}{m\left(v_0 + \dfrac{2eE_0}{m}t\right)}$$λ=m(v0​+m2eE0​​t)4mv0​λ0​​

Dividing the numerator and the denominator by $mv_0$mv0​, we get:

$$\lambda = \dfrac{4\lambda_0}{1 + \dfrac{2E_0 e}{m}\dfrac{t}{v_0}}$$λ=1+m2E0​e​v0​t​4λ0​​

Answer: $\dfrac{4\lambda_0}{\left[1 + \dfrac{2E_0 e}{m}\dfrac{t}{v_0}\right]}$[1+m2E0​e​v0​t​]4λ0​​