Two electrons are moving with non-relativistic speeds perpendicular to each other. If corresponding de Broglie wavelengths are λ_1 and λ_2 , their de Broglie wavelength in the frame of reference attached to their centre

Correct answer: C. λ_ C M = 2 λ_1 λ_2 λ_1^2+λ_2^2 Momentum (p) of each electron h λ_1 i and h λ_2 j Velocity of centre of mass V_ cm = h 2 ~m λ_1 i + h 2 ~m λ_2 j ( p=mv ) Velocity of 1 st particle about centre of mass V_ 1 ~cm = h 2 ~m λ_1 i - h 2 ~m λ_2 j λ_ cm = h h^2 4 λ_1^2 + h^2 4 λ_2^2 = 2 λ_1 λ_2 λ_1^2+λ_2^2 ( λ= h p )

JEE Main 201815 AprilMorning ShiftOnline

Exam: JEE Main
Year: 2018
Date: 15 April
Shift: Morning Shift
Mode: Online
Subject: Physics
Chapter: Atomic Physics

JEE Main 2018 · 15 April · Morning Shift · Online · Physics · Atomic Physics

Two electrons are moving with non-relativistic speeds perpendicular...

Two electrons are moving with non-relativistic speeds perpendicular to each other. If corresponding de Broglie wavelengths are

\lambda_1

and

\lambda_2

, their de Broglie wavelength in the frame of reference attached to their centre of mass is:

Options

A $\lambda_{\mathrm{CM}}=\lambda_1=\lambda_2$
B $\frac{1}{\lambda_1}=\frac{1}{\lambda_1}+\frac{1}{\lambda_2}$
C $\lambda_{C M}=\frac{2 \lambda_1 \lambda_2}{\sqrt{\lambda_1^2+\lambda_2^2}}$
D $\lambda_{C M}=\left(\frac{\lambda_1+\lambda_2}{2}\right)$

Correct answer

C. $\lambda_{C M}=\frac{2 \lambda_1 \lambda_2}{\sqrt{\lambda_1^2+\lambda_2^2}}$

Step-by-step solution

Momentum (p) of each electron

\frac{\mathrm{h}}{\lambda_1} \hat{\mathrm{i}}

and

\frac{\mathrm{h}}{\lambda_2} \hat{\mathrm{j}}

Velocity of centre of mass

\mathrm{V}_{\mathrm{cm}}=\frac{\mathrm{h}}{2 \mathrm{~m} \lambda_1} \hat{\mathrm{i}}+\frac{\mathrm{h}}{2 \mathrm{~m} \lambda_2} \hat{\mathrm{j}} \quad(\because $\mathrm{p}=\mathrm{mv}$)

Velocity of 1 st particle about centre of mass

V_{1 \mathrm{~cm}}=\frac{\mathrm{h}}{2 \mathrm{~m} \lambda_1} \hat{\mathrm{i}}-\frac{\mathrm{h}}{2 \mathrm{~m} \lambda_2} \hat{\mathrm{j}}

\lambda_{\mathrm{cm}}=\frac{\mathrm{h}}{\sqrt{\frac{\mathrm{h}^2}{4 \lambda_1^2}+\frac{\mathrm{h}^2}{4 \lambda_2^2}}}=\frac{2 \lambda_1 \lambda_2}{\sqrt{\lambda_1^2+\lambda_2^2}} \quad

\left(\because \lambda=\frac{\mathrm{h}}{\mathrm{p}}\right)

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