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

$2meV$

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

$\sqrt{meV}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

$\sqrt{2meV}$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D

$meV$

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is C $\sqrt{2meV}$Step 1: GivenPotential difference =$Vesu$charge of electron = $e$mass of electron = $m$Plank's constant, $h$ Step 2: Formula usedKinetic energy $=eV$ $eV=\frac{1}{2}m{u}^{2}\phantom{\rule{0ex}{0ex}}u=\sqrt{\frac{2eV}{m}}-----\left(1\right)$ Here, $u$ is velocity of electron.The de broglie wavelength, $\lambda =\frac{planck\text{'}scons\mathrm{tan}t}{momentum}=\frac{h}{mu}$$\frac{h}{\lambda }=mu-----\left(2\right)$ Step 3: CalculationSubstituting eq (1) in eq (2)$\frac{h}{\lambda }=m\sqrt{\frac{2eV}{m}}\phantom{\rule{0ex}{0ex}}\frac{h}{\lambda }=\sqrt{2meV}$Hence, option (c) is correct.

Suggest Corrections
0
Explore more