Question

An electron with speed v and a photon with a speed c have the same de-Brogli wavelength. If the K.E. and momentum of electrons is E_eand P_e and that of photon is E_ph and P_ph respectively, then the correct statement is –

• A. $\frac{E_{e}}{E_{ph}} = \frac{2c}{v}$
• B. $\frac{E_{e}}{E_{ph}} = \frac{v}{2c}$
• C. $\frac{P_{e}}{P_{ph}} = \frac{2c}{v}$
• D. None of these

Answer 1

Answer: (B)

$\frac{E_{e}}{E_{ph}} = \frac{v}{2c}$

Answer 2

$\frac{KE_{e}}{KE_{ph}}$ = $\frac{\frac{1}{2}mv^{2}}{h\nu}$= $\frac{1}{2}$× v × $\frac{mv}{h}$× $\frac{h}{h\nu}$

but $\frac{h}{mv}$ = $\frac{hc}{h\nu}$

$\frac{mv}{h} = \frac{h\nu}{hc}$

$\frac{E_{e}}{E_{ph}}$= $\frac { 1 } { 2 } \mathrm { v }$ $\left\lbrack \frac{mv}{h}.\frac{h}{h\nu}.\frac{c}{c} \right\rbrack$

= $\frac{1}{2}$v$\left\lbrack \frac{h\nu}{hc}.\frac{hc}{h\nu}.\frac{1}{c} \right\rbrack$ = $\frac{v}{2c}$