Question

An electron of mass $\mathrm{m}$ and photon has the same energy $\mathrm{E}$. The ratio of the wavelength of an electron to the photon: ($\mathrm{C}$ being velocity of light)

• A. ${\left(\frac{E}{2m}\right)}^{\frac{1}{2}}$
• B. $C{\left(2mE\right)}^{\frac{1}{2}}$
• C. $\frac{1}{c}{\left(\frac{E}{2m}\right)}^{\frac{1}{2}}$
• D. $\frac{1}{c}{\left(\frac{2m}{E}\right)}^{\frac{1}{2}}$

Answer 1

The correct option is C

$\frac{1}{c}{\left(\frac{E}{2m}\right)}^{\frac{1}{2}}$

The Explanation for the correct answer:

For electron,

De Broglie’s wavelength, ${\lambda }_e=\frac{h}{mv}$ $\left(\mathrm{p}=\mathrm{mv}\right)$

So, $\mathrm{p}=\sqrt{2\mathrm{mE}}$

Now, ${\lambda }_e=\frac{h}{\sqrt{2mE}}$……..1

For photon,

${\lambda }_P=\frac{hc}{E}$…………………2

From 1 and 2 we get

$\frac{{\mathrm{\lambda }}_{\mathrm{p}}}{{\mathrm{\lambda }}_{\mathrm{e}}}$$=\frac{\mathrm{hE}}{\sqrt{2\mathrm{mE}}\mathrm{hc}}$

$=\frac{1}{\mathrm{c}}{\left(\frac{\mathrm{E}}{2\mathrm{m}}\right)}^{\frac{1}{2}}$

So, option (C) is correct.