JEE Main 2018 — Physics Modern Physics

2018mcqmedium

Two electrons are moving with non-relativistic speeds perpendicular to each other. If corresponding de brogile wavelengths are ${\lambda }_{1}$ and ${\lambda }_{2}$ , their de brogile wavelength in the frame of reference attached to their centre of mass is:

Official previous-year question

Held on 15 Apr 2018 · Verified 6 Jul 2026.

Options

A
$\frac{1}{{\lambda }_{CM}}=\frac{1}{{\lambda }_{1}}+\frac{1}{{\lambda }_{2}}$
B
${\lambda }_{CM}=\frac{2{\lambda }_{1}{\lambda }_{2}}{\sqrt{{{\lambda }_{1}}^{2}+{{\lambda }_{2}}^{2}}}$
C
${\lambda }_{CM}={\lambda }_{1}={\lambda }_{2}$
D
${\lambda }_{CM}=\frac{{\lambda }_{1}+{\lambda }_{2}}{2}$

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Solution

Momentum of each electron $\frac{h}{{\lambda }_{1}} \hat{i} and \frac{h}{{\lambda }_{2}}\hat{j}$

Velocity of centre of mass

${V}_{cm}=\frac{h}{2m{\lambda }_{1}} \hat{i}+\frac{h}{2m{\lambda }_{2}}\hat{j}$

Velocity of ${1}^{st}$ particle about centre of mass

${V}_{cm}=\frac{h}{2m{\lambda }_{1}} \hat{i}-\frac{h}{2m{\lambda }_{2}}\hat{j}$

${\lambda }_{cm}=\frac{h}{\sqrt{\frac{{h}^{2}}{4{\lambda }_{1}^{2}}+\frac{{h}^{2}}{4{\lambda }_{2}^{2}}}}=\frac{2{\lambda }_{1} {\lambda }_{2}}{\sqrt{{\lambda }_{1}^{2}+{\lambda }_{2}^{2}}}$

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Options

$\frac{1}{{\lambda }_{CM}}=\frac{1}{{\lambda }_{1}}+\frac{1}{{\lambda }_{2}}$

${\lambda }_{CM}=\frac{2{\lambda }_{1}{\lambda }_{2}}{\sqrt{{{\lambda }_{1}}^{2}+{{\lambda }_{2}}^{2}}}$

${\lambda }_{CM}={\lambda }_{1}={\lambda }_{2}$

${\lambda }_{CM}=\frac{{\lambda }_{1}+{\lambda }_{2}}{2}$

Solution

Momentum of each electron $\frac{h}{{\lambda }_{1}} \hat{i} and \frac{h}{{\lambda }_{2}}\hat{j}$

Velocity of centre of mass

${V}_{cm}=\frac{h}{2m{\lambda }_{1}} \hat{i}+\frac{h}{2m{\lambda }_{2}}\hat{j}$

Velocity of ${1}^{st}$ particle about centre of mass

${V}_{cm}=\frac{h}{2m{\lambda }_{1}} \hat{i}-\frac{h}{2m{\lambda }_{2}}\hat{j}$

${\lambda }_{cm}=\frac{h}{\sqrt{\frac{{h}^{2}}{4{\lambda }_{1}^{2}}+\frac{{h}^{2}}{4{\lambda }_{2}^{2}}}}=\frac{2{\lambda }_{1} {\lambda }_{2}}{\sqrt{{\lambda }_{1}^{2}+{\lambda }_{2}^{2}}}$

JEE Main 2018 — Physics Modern Physics

Options

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Solution

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JEE Main Modern Physics in other years

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JEE Main 2018 — Physics Modern Physics

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