Question: The kinetic energy of hydrogen atom wave functions for which lis its minimum value of 0 is all radial. This is the case for the 1s and 2s states. The 2 p state has some rotational kinetic energy and some radial. Show that for very large n, the states of largest allowed lhave essentially no radial kinetic energy. Exercise 55 notes that the expectation value of the kinetic energy (including both rotational and radial) equals the magnitude of the total energy. Compare this magnitude with the rotational energy alone,$L^2/2m{r}^2$
,assuming that n is large. That lis as large as it can be, and that$r\equiv n^2{a}_0$.

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

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

${\epsilon }_0$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-

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

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

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

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

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

Solving further,

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

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.