Question

An electron and a photon possess the same de-Broglie wavelength. If $E_e$ and $ E_p$ respectively are the energies of electron and photon and v and c are their respective velocities, then $ \frac{ E_e }{ E_p} $ is equal to

• A. $ \frac{ v}{ c} $
• B. $ \frac{ v}{ 2c} $
• C. $ \frac{ v}{ 3c} $
• D. $ \frac{ v}{4 c} $

Answer 1

Answer: (B)

de-Broglie wavelength $ \lambda $
= $ \frac{ h }{ \sqrt{ 2 m E_e }} = \frac{ hc}{ E_p }$ or $2mE_e = \frac{ E_p^2 }{ c^2 } $ ... ( i)
But $ E_e = \frac{ 1}{ 2} mv^2$ or $ m = \frac{ 2 E_e }{ v^2 } $
From Eq. (i)
2 $ \bigg [ \frac{ 2 E_e }{ v^2 } \bigg ] E_e = \frac{ E_p^2 }{ c^2 } $
or $ \frac{ 4 E_e^2 }{ v^2 } = \frac{ E_p^2 }{ c^2 } $ or $\frac{ E_e^2 }{ E_p^2 } = \frac{ v^2 }{ 4c^2 } $
or $ \frac{ E_e }{ E_p } = \frac{ v}{ 2 c} $