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Question

A photon and electron have same de-Broglie wavelength. Give that v is the speed of electron and c is the velocity of light. Ee,Ep are the kinetic energy of electron and photon respectively. Pe,Ph are the momentum of electron and photon respectively. Then which of the following relation is correct?


A

EeEp=v2c

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B

EeEp=2cv

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C

PePh=c2v

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D

PePh=2cv

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Solution

The correct option is A

EeEp=v2c


As,
Ee=12mv2=12(mv)v=12(hλ)v
and Ep=hcλ
EeEp=v2c
Now, Pe=mv=hλ
and Ph=hλ
PePh=1

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