Question

If the kinetic energy of the moving particle is $E$, then the de Broglie wavelength is

• A. $\lambda =\frac{h}{\sqrt{2mE}}$
• B. $\lambda =\frac{\sqrt{2mE}}{h}$
• C. $\lambda =h\sqrt{2mE}$
• D. $\lambda =\frac{h}{E\sqrt{2m}}$

Answer 1

The correct option is A $\lambda =\frac{h}{\sqrt{2mE}}$

De Broglie wavelength of the particle, $\lambda =\frac{h}{p}$ where $p$ is the momentum of the particle

Kinetic energy of particle, $E=\frac{1}{2}m{v}^2=\frac{p^2}{2m}$ $(\because p=mv)$

Thus, we get: $p=\sqrt{2mE}$

$⟹$ $\lambda =\frac{h}{\sqrt{2mE}}$