Question

If photon energy $E$ and an electron have same energy $E$ (kinetic energy) and De-Broglie wavelength of an electron is ${\lambda }_e$ and De-Broglie wavelength of photon is ${\lambda }_p$. The correct relation between ${\lambda }_e$ and ${\lambda }_p$ is

• A. ${\lambda }_p\propto {\lambda }_e$
• B. ${\lambda }_p\propto \sqrt{{\lambda }_e}$
• C. ${\lambda }_p\propto \frac{1}{\sqrt{{\lambda }_e}}$
• D. ${\lambda }_p\propto {\lambda }_e^2$

Answer 1

The correct option is D ${\lambda }_p\propto {\lambda }_e^2$

For photon

$E=\frac{hc}{{\lambda }_p}$---------------$\left(1\right)$

$\Rightarrow {\lambda }_p=\frac{hc}{E}$

$E=\frac{1}{2}m{v}^2$

$\Rightarrow v=\sqrt{\frac{2E}{m}}$

${\lambda }_e=\frac{h}{mv}=\frac{h}{n\sqrt{\frac{2E}{m}}}=\frac{h}{\sqrt{2Em}}$

$\Rightarrow {{\lambda }_e}^2=\frac{h^2}{2Em}$

$\Rightarrow E=\frac{h^2}{2m{{\lambda }_e}^2}$

$\therefore \frac{h_e}{{\lambda }_p}=\frac{h^2}{2m{{\lambda }_e}^2}$

$\Rightarrow {\lambda }_p\alpha {{\lambda }_e}^2$

Hence,

option $\left(D\right)$ is correct answer.