Question

If wavelength of photon and electron is same then ratio of total energy of electron to total energy of photon would be :

• A. $\frac{Velocityofelectron}{Ligh{t}^{\prime }sspeed}$
• B. $\frac{Ligh{t}^{\prime }sspeed}{Electo{n}^{\prime }sspeed}$
• C. $\frac{Ligh{t}^{\prime }sspeed}{Velocityofelectron}$
• D. $\frac{Electro{n}^{\prime }sspeed}{Ligh{t}^{\prime }sspeed}$

Answer 1

Answer: (A)

$\frac{Velocityofelectron}{Ligh{t}^{\prime }sspeed}$

Given, ${\lambda }_e={\lambda }_{ph}$

where ${\lambda }_e$ and ${\lambda }_{ph}$ are wavelength of $e^{-}$ and photon respectively.

$\therefore E_{ph}=\frac{hc}{{\lambda }_{ph}}$

and whereas for $e^{-},{\lambda }_e=\frac{h}{p}=\frac{h}{mv}$

$\Rightarrow m=\frac{h}{{\lambda }_{e}v}$

$\therefore E_e=\frac{1}{2}\frac{h}{{\lambda }_{e}v}\times v^2=\frac{1}{2}\frac{hv}{{\lambda }_e}$

$\Rightarrow \frac{E_e}{E_{ph}}=\frac{1}{2}\frac{hv}{{\lambda }_e}\times \frac{{\lambda }_{ph}}{hc}=\frac{v}{2c}$