Question

A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference $Vesu$. If $e$ and $m$ are the charge and mass of an electron, respectively, then the value of $\frac{h}{\lambda }$ is given as?

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

Answer 1

The correct option is C

$\sqrt{2meV}$

Step 1: Given

Potential difference =$Vesu$

charge of electron = $e$

mass of electron = $m$

Plank's constant, $h$

Step 2: Formula used

Kinetic energy $=eV$

$$\begin{gathered} eV=\frac{1}{2}m{u}^2 \\ u=\sqrt{\frac{2eV}{m}}-----\left(1\right) \end{gathered}$$

Here, $u$ is velocity of electron.

The de broglie wavelength, $\lambda =\frac{planck\text{'}scons\tan t}{momentum}=\frac{h}{mu}$

$\frac{h}{\lambda }=mu-----\left(2\right)$

Step 3: Calculation

Substituting eq (1) in eq (2)

$$\begin{gathered} \frac{h}{\lambda }=m\sqrt{\frac{2eV}{m}} \\ \frac{h}{\lambda }=\sqrt{2meV} \end{gathered}$$

Hence, option (c) is correct.