Question

In experiment of Davisson-Germer, emitted electron from filament is accelerated through voltage V then de-Broglie wavelength of that electron will be ______m.

• A. $\frac{2 V e m}{\sqrt{h}}$
• B. $\frac{\sqrt{h}}{2 V e m}$
• C. $\frac{\sqrt{2 V e m}}{h}$
• D. $\frac{h}{\sqrt{2 V e m}}$

Answer 1

Answer: (D)

Kinetic energy $\frac{1}{2}mv^{2}=eV$
$v=\sqrt{\frac{2 e V}{m}}$
Wavelength $\lambda =\frac{h}{m v}$
$\lambda =\frac{h}{m \sqrt{\frac{2 e V}{m}}}=\frac{h}{\sqrt{2 e V m}}$