Question

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

Short answer

The relationship between the de Broglie wavelength and the kinetic energyis,

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

Step-by-step solution

Understanding de Broglie wavelength:

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.

Relationship between de Broglie's wavelength and Momentum:

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.

Relationship between Momentum and Kinetic Energy of the electron.

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.

Relationship between de Broglie's wavelength and Kinetic Energy of the electron.

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

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