Question

An electron (of mass $m$) and a photon have the same energy $E$ in the range of a few $\text{eV}$. The ratio of the de-Broglie wavelength associated with the electron and the wavelength of the photon is ($c=$ speed of light in vacuum)

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

Answer 1

The correct option is C $\frac{1}{c}{\left(\frac{E}{2m}\right)}^{1/2}$

de-Broglie wavelength of electron (${\lambda }_e$) is given by,

${\lambda }_e=\frac{h}{p_e}$

$\because p=\sqrt{2mE}$,

$\Rightarrow {\lambda }_e=\frac{h}{\sqrt{2mE}}\dots \left(i\right)$

Energy of photon $E$ is given by

$E=\frac{hc}{{\lambda }_p}$
Here ${\lambda }_p=$ wavelength of Photon

$\Rightarrow {\lambda }_p=\frac{hc}{E}\dots \left(ii\right)$

On dividing $\left(i\right)$ by $\left(ii\right)$, we get,

$\Rightarrow \frac{{\lambda }_e}{{\lambda }_p}=\frac{h}{\sqrt{2mE}}\cdot \frac{E}{hc}=\frac{1}{c}\cdot \sqrt{\frac{E}{2m}}$

Hence, option $\left(C\right)$ is correct.