Question

A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference of $V\text{esu}$. If $e$ and $m$ are charge and mass of an electron, respectively, then the value of $\frac{h}{\lambda }$ (where, $\lambda$ is wavelength associated with electron) is given by:

• A. $2meV$
• B. $\sqrt{meV}$
• C. $\sqrt{2meV}$
• D. $meV$

Answer 1

The correct option is C $\sqrt{2meV}$

De Broglie wavelenth for an electron $\lambda =\frac{h}{p}$
$\Rightarrow p=\frac{h}{\lambda }...\left(i\right)$
Kinetic energy of an electron $=eV$
As we know that,
$KE=\frac{p^2}{2m}$
$\therefore eV=\frac{p^2}{2m}$
$\Rightarrow p=\sqrt{2meV}...\left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$
$\frac{h}{\lambda }=\sqrt{2meV}$