Here we investigate the link between nand l, reflected in equation (7-33). (a) Show that if a classical point charge were held in a circular orbit about a fixed point charge by the Coulomb force, its kinetic energy would be given by $KE=e^2/8\pi {\epsilon }_{0}r$ (b) According to equation (7-30), the rotational kinetic energy in hydrogen is $h^{2}l(l+1)/2m{r}^2$. Of course, ris not well defined for a “cloud”, but by using_{$r=n^2{a}_0$}argue that the condition that l not exceed n is reasonable.

The azimuthal quantum number specifies the angular momentum as well as the shape of the atomic orbital.

$\left|L\right|\mathbf{=}\sqrt{\mathbf{l}\left(l+1\right)}\mathbf{\times }\mathbf{h}\mathbf{;}\mathit{l}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{3}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(n-1\right)$$\left|L\right|=\sqrt{l(l+1)}\times h;I=0,1,2,3..............(n-1)$

Where, nis the principal quantum number, Iis the Azimuthal quantum number, his Planck’s constant, Lis the Orbital angular momentum.

As you know that, Coulomb force is,
$F=\frac{q^2}{4\pi {\epsilon }_0{r}^2}$

Where, q is the charge on the objects, r is the distance,${\epsilon }_0$

is the Permitivity of free space.

The acceleration of the charge is given by,

$a=\frac{v^2}{r}$

Where, v is the speed of the object and r is the radius of curvature

Write the equation for Newton’s second law of motion as given below.

F=ma

Where, F is the force, m is the mass, and a is the acceleration.

$\begin{array}{rcl}\frac{e^2}{4\pi {\epsilon }_0{r}^2}& =& m\frac{v^2}{r}\\ \frac{e^2}{8\pi {\epsilon }_{0}r}& =& \frac{1}{2}m{v}^2\end{array}$

Where, e is the charge on an electron.

The kinetic energy is

$$\begin{gathered} KE=\frac{1}{2}m{v}^2 \\ KE=\frac{e^2}{8\pi {\epsilon }_{0}r}......\left(1\right) \end{gathered}$$

Rotational energy of the particle is,

$E=\frac{l(l+1)h^2}{2m{r}^2}$

Where, h is Planck’s constant.

$$\begin{gathered} E=\frac{I(I+1)h^2}{2mr}\frac{1}{a_0}\frac{1}{n^2} \\ =\frac{I(I+1)h^2}{2mr}\frac{m{e}^2}{4\pi {\epsilon }_0{h}^2}\frac{1}{n^2} \\ E=\frac{e^2}{2\left(4\pi {\epsilon }_o\right)r}\frac{I(I+1)}{n^2} \end{gathered}$$

From equations (1) and (2), you get, if I exceeds n, the rotational energy would exceed kinetic energy in circular orbit and the sum of the energies will be more than the ‘total energy’.

Hence, the argue that the condition thatI not exceed n is reasonable.