Question

An electron of mass $\mathrm{m}$ and $\mathrm{a}$ photon have the same energy $\mathrm{E}$. The ratio of the wavelength of an electron to that of the photon is : ($\mathrm{C}$ being the 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. $\frac{1}{c}\left(\frac{2m}{E}\right)$

Answer 1

The correct option is C

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

The Explanation for the correct option:

De Broglie’s wavelength ${\lambda }_e=\frac{h}{mv}$

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

Now,

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

For a photon,${\lambda }_P=\frac{hc}{E}$………………$2$

From 1 and 2 we get,

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

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

Hence,Option (C) is correct.