Question

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

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

Answer 1

Answer: (C)

Kinetic energy of the moving particle is given by, $E=\frac{1}{2}m{v}^2$
$\Rightarrow mv=\sqrt{2mE}...\left(1\right)$
where, $mv$ is the momentum of a moving particle.
de-Broglie wavelength of a moving particle is given by, $\lambda =\frac{h}{mv}...\left(2\right)$
where, $h$ is a Planck's constant.
Using equation $\left(1\right)$ and $\left(2\right)$ ,
$\Rightarrow \lambda =\frac{h}{\sqrt{2mE}}$