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 reference attached to their centre of mass is:
1λCM=1λ1+1λ2
λCM=λ1=λ2
λCM=2λ1λ2√λ12+λ22
λCM=(λ1+λ22)
A
λCM=2λ1λ2√λ12+λ22
B
λCM=λ1=λ2
C
1λCM=1λ1+1λ2
D
λCM=(λ1+λ22)
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Solution
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Since λ=hmv
v=hmλ
Let v1 and v2 are the speeds of electrons
vcm=v1+v22
h2mλcm=12(hmλ1+hmλ2)
1λcm=1λ1+1λ2
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