91影视

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{脳}\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)}脳h;I=0,1,2,3..............(n-1)$

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

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

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

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鈥檚 second law of motion as given below.

F=ma

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

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

Where, e is the charge on an electron.

The kinetic energy is

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

Rotational energy of the particle is,

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

Where, h is Planck鈥檚 constant.

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

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 鈥榯otal energy鈥�.

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