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

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

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 $\left(\mathrm{\lambda }\right)$, and the wavelength depends on its momentum $\left(\mathrm{p}\right)$; hence you can write,

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

Here, h is called the Plancks constant, and $\mathrm{\lambda }$is called the de Broglie's wavelength.

Relationship between Momentum and Kinetic Energy of the electron.

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

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

….. (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

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