A photon has the same wavelength as the de-Broglie wavelength of electron. Given that c = speed of light and v = speed of electron. Which of the following relation is correct? [Here $E_e$ =Kinetic energy of electron, $E_{ph}$ = energy of photon, $p_e$ =momentum of electron and $p_{ph}$ =momentum of photon]
Answer & explanation
Correct answer: option 2
Given,
$λ_{ph}=λ_e$
$∴\frac{h}{p_{ph}}=\frac{h}{p_e}$
$∴p_{ph}=p_e$
$∴\frac{E_{ph}}{c}=\frac{2E_e}{v}$
$∴\frac{E_e}{E_{ph}}=\frac{v}{2c}$