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 electron are $E_{e}$ and $p _{ e }$ and that of photon are $E _{ ph }$ and $p _{ ph }$ respectively. Which of the following is correct?

• A. $\frac{E_{e}}{E_{p h}}=\frac{2 c}{v}$
• B. $\frac{E_{e}}{E_{p h}}=\frac{v}{2 c}$
• C. $\frac{ p _{ e }}{ p _{ ph }}=\frac{2 c }{ v }$
• D. $\frac{ p _{ e }}{ p _{ ph }}=\frac{ v }{2 c }$

Answer 1

Answer: (B)

$\lambda_{ e }=\lambda_{\text {photon }}$
$\frac{ h }{ mv }=\frac{ h }{ P _{\text {photon }}}$
$\Rightarrow P _{\text {photon }}= mv$
$\frac{ E _{ c }}{ E _{ ph }}=\frac{\frac{1}{2} mv ^{2}}{\frac{ hc }{\lambda}}=\frac{1}{2} \frac{ mv }{ P _{ Ph } C } \times v =\frac{ v }{2 C }$