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JEE PHYSICS MCQ: DUAL NATURE OF RADIATION AND MATTER

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Analyzing +4 | -1
A photon has same wavelength as the de-Broglie wavelength of electrons. Given C=C= Speed of light V=\mathrm{V}= Speed of electron

Ee=E_e= Kinetic energy of electrons

EPh=E_{P h}= Kinetic energy of photon

Pe=P_e= Momentum of electron

PPh=P_{Ph}= Momentum of photon

Which of the following is correct?
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Correct option: B
EeEPh=V2C\frac{E_e}{E_{{Ph }}}=\frac{V}{2 C}
Option B (Ee/EPh = V/2C) is correct because it correctly relates the kinetic energies of the electron and photon based on their equal wavelengths.

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