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 referenceattached to their centre of mass is : [Main 15 April 2018 S1]

Correct Answer is : λCM=2λ1λ2√λ12+λ22

\lambda_{CM}=\lambda_1=\lambda_2

\lambda_{CM}={2\lambda_1\lambda_2}/\sqrt{\lambda_1^2+\lambda_2^2}

1/\lambda_{CM}=1/\lambda_1+1/\lambda_2

1/\lambda_{CM}=({\lambda_1+\lambda_2}/2)

JEE Main 15 Apr 2018 Shift 1 Solved Paper

Section: Physics

Question : 24 of 90

Marks: +1, -0

Two electrons are moving with non-relativistic speeds perpendicular to each other. If corresponding de Broglie wavelengths are λ1 and λ2, their de Broglie wavelength in the frame of referenceattached to their centre of mass is :

[Main 15 April 2018 S1]

λCM=λ1=λ2
λCM=
2λ1λ2
√λ12+λ22
1
λCM
=
1
λ1
+
1
λ2
1
λCM
=(
λ1+λ2
2
)

Validate

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