An electron (mass m ) with… - JEE Main 2020 Physics PYQ

JEE Main 2020 — Physics Modern Physics

2020mcqhard

An electron (mass $m$ ) with initial velocity $\vec{v}={v}_{0}\hat{i}+{v}_{0}\hat{j}$ is in an electric filed $\vec{E}=-{E}_{0}\hat{k}.$ If ${\lambda }_{0}$ is initial de-Broglie wavelength of electron, its de-Broglie wave length at time $t$ is given by:

Official previous-year question

Held on 8 Jan 2020 · Verified 6 Jul 2026.

Options

A
$\frac{{\lambda }_{0}\sqrt{2}}{\sqrt{1+\frac{{e}^{2}{E}^{2}{t}^{2}}{{m}^{2}{v}_{0}^{2}}}}$
B
$\frac{{\lambda }_{0}}{\sqrt{1+\frac{{e}^{2}{E}^{2}{t}^{2}}{{m}^{2}{v}_{0}^{2}}}}$
C
$\frac{{\lambda }_{0}}{\sqrt{1+\frac{{e}^{2}{E}^{2}{t}^{2}}{2{m}^{2}{v}_{0}^{2}}}}$
D
$\frac{{\lambda }_{0}}{\sqrt{2+\frac{{e}^{2}{E}^{2}{t}^{2}}{{m}^{2}{v}_{0}^{2}}}}$

Did you get this right?

Solution

Initial velocity of electron,

$v={v}_{0}\hat{i}+{v}_{0}\hat{j}\Rightarrow |u|=\sqrt{{v}_{0}^{2}+{v}_{0}^{2}}={v}_{0}\sqrt{2}$

Initial de Broglie wavelength associated with electron,

${\lambda }_{0}=\frac{h}{p}=\frac{h}{m{v}_{0}\sqrt{2}}$

now,

$v=u+at$

Velocity as a function of time $={v}_{0}\hat{i}+{v}_{0}\hat{j}+\frac{e{E}_{0}}{m}t\hat{k}$

so wavelength $\lambda =\frac{h}{m\sqrt{2{v}_{0}^{2}+\frac{{e}^{2}{E}_{0}^{2}}{{m}^{2}}{t}^{2}}}$

$\lambda =\frac{{\lambda }_{0}}{\sqrt{1+\frac{{e}^{2}{E}_{0}^{2}}{2{m}^{2}{v}_{0}^{2}}{t}^{2}}}$

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Options

$\frac{{\lambda }_{0}\sqrt{2}}{\sqrt{1+\frac{{e}^{2}{E}^{2}{t}^{2}}{{m}^{2}{v}_{0}^{2}}}}$

$\frac{{\lambda }_{0}}{\sqrt{1+\frac{{e}^{2}{E}^{2}{t}^{2}}{{m}^{2}{v}_{0}^{2}}}}$

$\frac{{\lambda }_{0}}{\sqrt{1+\frac{{e}^{2}{E}^{2}{t}^{2}}{2{m}^{2}{v}_{0}^{2}}}}$

$\frac{{\lambda }_{0}}{\sqrt{2+\frac{{e}^{2}{E}^{2}{t}^{2}}{{m}^{2}{v}_{0}^{2}}}}$

Solution

Initial velocity of electron,

$v={v}_{0}\hat{i}+{v}_{0}\hat{j}\Rightarrow |u|=\sqrt{{v}_{0}^{2}+{v}_{0}^{2}}={v}_{0}\sqrt{2}$

Initial de Broglie wavelength associated with electron,

${\lambda }_{0}=\frac{h}{p}=\frac{h}{m{v}_{0}\sqrt{2}}$

now,

$v=u+at$

Velocity as a function of time $={v}_{0}\hat{i}+{v}_{0}\hat{j}+\frac{e{E}_{0}}{m}t\hat{k}$

so wavelength $\lambda =\frac{h}{m\sqrt{2{v}_{0}^{2}+\frac{{e}^{2}{E}_{0}^{2}}{{m}^{2}}{t}^{2}}}$

$\lambda =\frac{{\lambda }_{0}}{\sqrt{1+\frac{{e}^{2}{E}_{0}^{2}}{2{m}^{2}{v}_{0}^{2}}{t}^{2}}}$

Options

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Solution

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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