Question

The wavelength of an electron $\left({\lambda }_e\right)$ and that of a photon $\left({\lambda }_p\right)$ of the same energy $E$, are related by,

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

Answer 1

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

For electron,

${\lambda }_e=\frac{h}{p}=\frac{h}{\sqrt{2mE}}$

$\Rightarrow E=\frac{h^2}{2m{\lambda }_e^2}\dots \left(1\right)$

For the photon,

$E=\frac{hc}{{\lambda }_p}\dots \left(2\right)$

As both electron and photon have same energies, equating $\left(1\right)$ and $\left(2\right)$, we get,

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

${\lambda }_e^2=\frac{h{\lambda }_p}{2mc}$

$\therefore$ we can write, ${\lambda }_e^2\propto {\lambda }_p$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.