Question

The kinetic energy of an electron in an orbit of radius $\mathrm{r}$in a hydrogen atom is ($\mathrm{e}$ is the electronic charge)?

Answer 1

Step 1: Given data:

The radius of the orbit of an electron in a hydrogen atom is given as = $\mathrm{r}$

We can calculate the kinetic energy of the electron, by using the postulates of Bohr theory.

Step 2: Formula for the kinetic energy:

The formula used for calculating the kinetic energy of the electron in a Hydrogen atom is given as:

$T=\frac{1}{2}m{v}^2$

Where $\mathrm{T}=$ Kinetic energy of the electron

$\mathrm{v}=$The velocity of the electron

Step 3: Calculating the value of velocity ($\mathrm{v}$):

To calculate the K.E of electrons in an orbit of a hydrogen atom, we have to use Bohr’s atomic model, which states that the electrons revolve around the nucleus in a circular orbit i.e stationary states, under the influence of Coulomb force.
Corresponding to each of the stationary states, the orbital angular momentum of the electron $mvr$ is equal to an integral multiple of $\hslash$, given as:

$$\begin{gathered} mvr=n\hslash \\ \Rightarrow v={\displaystyle \frac{n\hslash }{mr}} \end{gathered}$$
To maintain the stability of the circular orbit of the electrons, the Coulomb’s force of attraction is balanced by the centripetal force, given as:
${\displaystyle \frac{m{v}^2}{r}}={\displaystyle \frac{1}{4\pi {\epsilon }_0}}{\displaystyle \frac{e^2}{r^2}}$
$\Rightarrow v^2={\displaystyle \frac{e^2}{4\pi {\epsilon }_{0}mr}}$

Step 4: Calculating the value of kinetic energy:
Substituting the value of $v^2$ in this expression of kinetic energy we can get,
$$\begin{gathered} \mathrm{T}=\frac{1}{2}\mathrm{m}\times \frac{{\mathrm{e}}^2}{4\mathrm{\pi }{\mathrm{\epsilon }}_0\mathrm{mr}} \\ \Rightarrow \mathrm{T}=\frac{1}{4\mathrm{\pi }{\mathrm{\epsilon }}_0}\frac{{\mathrm{e}}^2}{2\mathrm{r}} \end{gathered}$$
Hence, the kinetic energy of an electron in an orbit of radius $\mathrm{r}$in a hydrogen atom is $\frac{1}{4\mathrm{\pi \epsilon }0}\frac{{\mathrm{e}}^2}{2\mathrm{r}}$.