Question: For an electron, apply the relationship between the de Broglie wavelength and the kinetic energy.

Wavelength in physics may be defined as the distance of one crest to another crest of a wave. Now according to Louis de Broglie, every particle follows wave nature.

An electron has a wavelength $\left(\lambda \right)$, and the wavelength depends on its momentum $\left(p\right)$; hence you can write,

$\lambda =\frac{h}{p}$ ..... (1)

Here, $h$is called Planck's constant, and $\lambda$is called the de Broglie's wavelength.

You know that as a body's kinetic energy $\left(K\right)$increases, it results in the rise of momentum $\left(p\right)$. Therefore, the kinetic energy of an electron directly depends on Momentum. Hence,

$$\begin{gathered} k=\frac{p^2}{2m} \\ {p}^2=2mk \end{gathered}$$

$p=\sqrt{2mk}$ ..... (2)

Here,$m$ is the mass of the electron.

By putting the value of equation (2) in equation (1), you get

$\lambda =\frac{h}{\sqrt{2mk}}$