Question

An electron of mass $m$ and a photon have same energy $E$. The ratio of de-Broglie wavelength associated with them is:

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

Answer 1

Answer: (A)

For an electron, ${\lambda }_e=\frac{h}{\sqrt{2mE}}$

For photon, $E=pc\Rightarrow {\lambda }_{Photon}=\frac{hc}{E}$

$\Rightarrow \frac{{\lambda }_e}{{\lambda }_{Photon}}=\frac{h}{\sqrt{2mE}}\times \frac{E}{hc}={\left(\frac{E}{2m}\right)}^{1/2}\frac{1}{c}$