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The equation for the total energy of a hydrogen atom is
$E_n=-\frac{m_e^4}{2{\left(4蟺{蔚}_0\right)}^2{h}^2}\frac{1}{n^2}$
$E_n=-\frac{m_e^4}{2{\left(4蟺{蔚}_0\right)}^2{h}^2}\frac{1}{n^2}$

Here,$E_n$is the total energy of the hydrogen atom,

m is the mass of the particle,

e is the charge of the electron

${蔚}_0$is the permittivity of free space

h the modified Planck's constant, and

n is the principal quantum number

The average value of the radius of an orbit of the electron in the ground state, around the nuclei of a hydrogen atom, is known as the Bohr radius.

The magnitude of the total energy of a hydrogen atom is-

$\left|E_n\right|=-\frac{m{e}^4}{2{\left(4蟺{蔚}_0\right)}^2{h}^2}\frac{1}{n^2}......\left(1\right)$

The angular momentum for a particle is given as-

$L=\sqrt{l(l+1)h}$

Where, L is the angular momentum of a particle and l is the azimuthal quantum number

The expression for Bohr's radius is

$a_0=\frac{\left(4蟺{蔚}_0\right)h^2}{m{e}^2}$

Where$a_0$is the Bohr radius.

Here, electron is carrying two type of kinetic energy- Rotational and Radial. So, total kinetic energy will be the sum of both .

The expression for rotational kinetic energy is
$E_{rot}=\frac{L^2}{2m{r}^2}$

Where, $E_{rot}$is the rotational kinetic energy r is the radius of hydrogen atom.

Substitute $\sqrt{l(l+1)}h$for L and $n^2{a}_0$for r in $E_{rot}=\frac{L^2}{2m{r}^2}$

$$\begin{gathered} E_{rot}=\frac{{\left(\sqrt{l(l+1)}h\right)}^2}{2m{\left(n^2{a}_0\right)}^2} \\ =\frac{l(l+1)h^2}{2m{n}^4{a}_0^2} \end{gathered}$$

The range of l is from 0 to n - 1 . For maximum n , the corresponding value of maximum l is n - 1 .

Substitute n - 1 for l in $E_{rot}=\frac{l(l+1)h^2}{2m{n}^4{a}_0^2}$
$E_{rot}=\frac{(n-1)(n-1+1)h^2}{2m{n}^4{a}_0^2}$
$=\frac{(n-1)n{h}^2}{2m{n}^4{a}_0^2}$
$$\begin{gathered} 鈮�\frac{n^2{h}^2}{2m{n}^4{a}_0^2} \\ =\frac{h^2}{2m{n}^2{a}_0^2} \end{gathered}$$

Substitute $\frac{\left(4蟺{蔚}_0\right)h^2}{m{e}^2}$for$a_0$in $E_{rot}=\frac{h^2}{2m{n}^2{a}_0^2}$
$$\begin{gathered} E_{rot}=\frac{h^2}{2m{n}^2{\left({\displaystyle \frac{\left(4蟺{蔚}_0\right)h^2}{m{e}^2}}\right)}^2} \\ =\frac{h^2}{2m{n}^2\left({\displaystyle \frac{{\left(4蟺{蔚}_0\right)}^2{h}^4}{m^2{e}^4}}\right)} \\ =\frac{m^2{e}^4{h}^2}{2m{n}^2{\left(4蟺{蔚}_0\right)}^2{h}^2} \\ =\frac{m{e}^4}{2n^2{\left(4蟺{蔚}_0\right)}^2{h}^2} \end{gathered}$$

Solving further,

$E_{rot}=\frac{m{e}^4}{2{\left(4蟺{蔚}_0\right)}^2{h}^2}\frac{1}{n^2}.................\left(2\right)$

From the equation (1) and (2) the rotational kinetic energy of hydrogen atom is equal to the magnitude of total energy of hydrogen atom. Since for larger n the rotational kinetic energy of atom is equal to the total energy, that means there is no radial kinetic energy.

Therefore, for larger n, there is no radial kinetic energy.

On comparing equations (1) and (2) the magnitude of total energy is equal to the rotational kinetic energy.