An electron of mass m is moving… - JEE Main 2026 Physics PYQ

JEE Main 2026 — Physics Modern Physics

2026mcqmedium

An electron of mass $m$ is moving in an electric field $\vec{E} = -2E_0\hat{i}$ ( $E_0 =$ constant $> 0$ ), with an initial velocity $\vec{V} = v_0\hat{i}$ ( $v_0 =$ constant $> 0$ ). If $\lambda_0 = \dfrac{h}{4mv_0}$ , its de Broglie wavelength at time $t$ is __________.

( $e =$ charge of electron)

Official previous-year question

Held on 5 Apr 2026 · Verified 6 Jul 2026.

Options

A
$\dfrac{4\lambda_0}{\left[1 - \dfrac{E_0 e}{2m}\dfrac{t}{v_0}\right]}$
B
$\dfrac{4\lambda_0}{\left[1 + \dfrac{E_0 e}{2m}\dfrac{t}{v_0}\right]}$
C
$\dfrac{4\lambda_0}{\left[1 + \dfrac{2E_0 e}{m}\dfrac{t}{v_0}\right]}$
D
$\dfrac{4\lambda_0}{\left[1 - \dfrac{2E_0 e}{m}\dfrac{t}{v_0}\right]}$

Did you get this right?

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:

$\vec{F} = (-e)(-2E_0\hat{i}) = 2eE_0\hat{i}$

The acceleration of the electron is:

$\vec{a} = \dfrac{\vec{F}}{m} = \dfrac{2eE_0}{m}\hat{i}$

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

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

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

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

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

$\lambda = \dfrac{4mv_0\lambda_0}{m\left(v_0 + \dfrac{2eE_0}{m}t\right)}$

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

$\lambda = \dfrac{4\lambda_0}{1 + \dfrac{2E_0 e}{m}\dfrac{t}{v_0}}$

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

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Options

$\dfrac{4\lambda_0}{\left[1 - \dfrac{E_0 e}{2m}\dfrac{t}{v_0}\right]}$

$\dfrac{4\lambda_0}{\left[1 + \dfrac{E_0 e}{2m}\dfrac{t}{v_0}\right]}$

$\dfrac{4\lambda_0}{\left[1 + \dfrac{2E_0 e}{m}\dfrac{t}{v_0}\right]}$

$\dfrac{4\lambda_0}{\left[1 - \dfrac{2E_0 e}{m}\dfrac{t}{v_0}\right]}$

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:

$\vec{F} = (-e)(-2E_0\hat{i}) = 2eE_0\hat{i}$

The acceleration of the electron is:

$\vec{a} = \dfrac{\vec{F}}{m} = \dfrac{2eE_0}{m}\hat{i}$

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

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

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

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

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

$\lambda = \dfrac{4mv_0\lambda_0}{m\left(v_0 + \dfrac{2eE_0}{m}t\right)}$

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

$\lambda = \dfrac{4\lambda_0}{1 + \dfrac{2E_0 e}{m}\dfrac{t}{v_0}}$

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

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Options

Did you get this right?

Solution

Your note

JEE Main Modern Physics in other years

More JEE Main Modern Physics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Modern Physics