Question

In experiment of Davisson-Germer, emitted electron from filament is accelerated throguh voltage $V$ then de-Broglie wavelength of that electron will be ________ m .

• A. $\frac{2eVm}{\sqrt{h}}$
• B. $\frac{\sqrt{h}}{2Vem}$
• C. $\frac{\sqrt{2Vem}}{h}$
• D. $\frac{h}{\sqrt{2Vem}}$

Answer 1

The correct option is D $\frac{h}{\sqrt{2Vem}}$

Kinetic energy of the electron will be due to the voltage supplied.

$\frac{1}{2}m{v}^2=eV$
$v=\sqrt{\frac{2eV}{m}}$

We know that the wavelength is obtained as:

$\lambda =\frac{h}{mv}$
$\lambda =\frac{h}{m\sqrt{\frac{2eV}{m}}}=\frac{h}{\sqrt{2eVm}}$