An electron of mass ' m ' with an initial velocity \overrightarrowv=v_0 \hati(v_0\gt 0) enters an electric field \overrightarrowE=-E_o \hatk . If the initial de Broglie wavelength is \lambda_0 , the value after time t would be

\frac\lambda_o\sqrt1+\frace^2 E_o^2 t^2~m^2 v_o^2

\frac\lambda_o\sqrt1-\frace^2 E_o^2 t^2~m^2 v_o^2

\lambda_o \sqrt1+\frace^2 E_o^2 t^2~m^2 v_o^2

Dual Nature of Radiation PYQ — JEE Main 2025 · 24 Jan · Shift 1 · Q66 with Solution

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Correct answer: C

Initial momentum and de Broglie wavelength

For a particle, de Broglie wavelength is

\lambda = \frac{h}{p}

Initially, the electron moves with velocity

\vec v = v_0 \hat i

So initial momentum is

p_0 = mv_0

Hence,

\lambda_0 = \frac{h}{mv_0}

Force on the electron in the electric field

Given electric field:

\vec E = -E_0 \hat k

Charge of electron is

q = -e

Therefore force on electron:

\vec F = q\vec E = (-e)(-E_0\hat k)= eE_0\hat k

So acceleration is

\vec a = \frac{\vec F}{m} = \frac{eE_0}{m}\hat k

Velocity after time $t$

There is no force in the $x$ -direction, so

v_x = v_0

In the $z$ -direction,

v_z = at = \frac{eE_0}{m}t

Thus the velocity vector after time $t$ is

\vec v = v_0\hat i + \frac{eE_0 t}{m}\hat k

Its magnitude is

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

Momentum after time $t$

Magnitude of momentum:

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

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

New de Broglie wavelength

Using

\lambda = \frac{h}{p}

we get

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

Now since

\lambda_0 = \frac{h}{mv_0}

we write

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

Checking options

A has minus sign inside root, incorrect.
B says wavelength remains unchanged, incorrect because speed increases.
C matches exactly:

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

D is inverse of the correct result, incorrect.

Therefore, the correct option is C.

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View written solutionFree

Correct answer: C

Initial momentum and de Broglie wavelength

For a particle, de Broglie wavelength is

\lambda = \frac{h}{p}

Initially, the electron moves with velocity

\vec v = v_0 \hat i

So initial momentum is

p_0 = mv_0

Hence,

\lambda_0 = \frac{h}{mv_0}

Force on the electron in the electric field

Given electric field:

\vec E = -E_0 \hat k

Charge of electron is

q = -e

Therefore force on electron:

\vec F = q\vec E = (-e)(-E_0\hat k)= eE_0\hat k

So acceleration is

\vec a = \frac{\vec F}{m} = \frac{eE_0}{m}\hat k

Velocity after time $t$

There is no force in the $x$ -direction, so

v_x = v_0

In the $z$ -direction,

v_z = at = \frac{eE_0}{m}t

Thus the velocity vector after time $t$ is

\vec v = v_0\hat i + \frac{eE_0 t}{m}\hat k

Its magnitude is

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

Momentum after time $t$

Magnitude of momentum:

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

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

New de Broglie wavelength

Using

\lambda = \frac{h}{p}

we get

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

Now since

\lambda_0 = \frac{h}{mv_0}

we write

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

Checking options

A has minus sign inside root, incorrect.
B says wavelength remains unchanged, incorrect because speed increases.
C matches exactly:

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

D is inverse of the correct result, incorrect.

Therefore, the correct option is C.

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