Question

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

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

Answer 1

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

Given:
Charge of electron = e
Potential difference applied = $V$
Mass of electron = m

Let $v$ be the veloctiy of the electron. Kinetic energy would be =$\frac{1}{2}m{v}^2$

Kinetic energy gained by electron due to the applied potential
$=eV$
$\therefore eV=\frac{1}{2}m{v}^2⟹meV=\frac{1}{2}{m}^2{v}^2⟹2meV=(mv{)}^2⟹\sqrt{2meV}=mv$

de Broglie wavelength of the electron given by:
$\lambda =\frac{h}{mv}⟹\frac{h}{\lambda }=mv$
Since
$mv=\sqrt{2meV}$
$\frac{h}{\lambda }=\sqrt{2meV}$