Question

An electron with speed v and a photon with speed c have the same de-broglie wavelength. If the kinetic energy and momentum of electrons are E_e and P_e and that of photon are E_Ph and P_Ph respectively, then correct statement is –

• A. E_e/E_Ph= 2c/v
• B. E_e/E_Ph = v/2c
• C. P_e/P_Ph = 2c/v
• D. P_e/P_Ph = v/2c

Answer 1

Answer: (B)

E_e/E_Ph = v/2c

Answer 2

$\frac{(K.E.)_{e}}{(K.E.)_{Ph}} = \frac{1/2mv^{2}}{h\nu}$ = $\frac{1}{2} \times v \times \frac{mv}{h} \times \frac{h}{h\nu}$ but $\frac{h}{mv} = \frac{hc}{h\nu}$

\ $\frac { \mathrm { mv } } { \mathrm { h } } = \frac { \mathrm { h } v } { \mathrm { hc } }$ $\frac{E_{e}}{E_{Ph}} = \frac{1}{2}v$ $\left\lbrack \frac{mv}{h} \times \frac{h}{h\nu} \times \frac{c}{c} \right\rbrack$ = $\frac{1}{2}v\left\lbrack \frac{h\nu}{hc} \cdot \frac{hc}{h\nu} \cdot \frac{1}{c} \right\rbrack$

= v/2c