We have noted that for a given energy, as lincreases, the motion is more like a circle at a constant radius, with the rotational energy increasing as the radial energy correspondingly decreases. But is the radial kinetic energy 0 for the largest lvalues? Calculate the ratio of expectation values, radial energy to rotational energy, for the$\mathbf{(}\mathit{n}\mathbf{,}\mathit{l}\mathbf{,}{\mathit{m}}_{\mathbf{t}}\mathbf{)}\mathbf{=}\left(2.1,+1\right)$state. Use the operators

$$\begin{gathered} {\overbrace{\mathbf{K}\mathbf{E}}}_{\mathbf{r}\mathbf{a}\mathbf{d}}\mathbf{=}\frac{\mathbf{-}{\mathbf{h}}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}}\frac{\mathbf{1}}{{\mathbf{r}}^{\mathbf{2}}}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{r}}\left(r\frac{\partial }{\partial r}\right) \\ {\overbrace{\mathbf{K}\mathbf{E}}}_{\mathbf{r}\mathbf{a}\mathbf{d}}\mathbf{=}\frac{{\mathbf{h}}^{\mathbf{2}}\mathbf{l}\mathbf{(}\mathbf{l}\mathbf{+}\mathbf{1}\mathbf{)}}{\mathbf{2}\mathbf{m}{\mathbf{r}}^{\mathbf{2}}} \end{gathered}$$

Which we deduce from equation (7-30).

radial kinetic energy,${\overbrace{KE}}_{rad}=\frac{-h^2}{2m}\frac{1}{r^2}\frac{\partial }{\partial r}\left(r\frac{\partial }{\partial r}\right)$

rotational kinetic energy,${\overbrace{KE}}_{rad}=\frac{h^{2}l(l+1)}{2m{r}^2}$

Where, l is the orbital quantum number.

Expression for radial function having a principal quantum number equal to 2 and an orbital quantum number equal to 1 will be:

${\mathbf{R}}_{\mathbf{2}\mathbf{,}\mathbf{1}}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{\left(}\mathbf{2}{\mathbf{a}}_{\mathbf{0}}\mathbf{\right)}}^{\mathbf{3}\mathbf{/}\mathbf{2}}}\frac{\mathbf{r}}{\sqrt{\mathbf{3}}{\mathbf{a}}_{\mathbf{0}}}{\mathbf{e}}^{\mathbf{-}\mathbf{r}\mathbf{\left(}\mathbf{2}{\mathbf{a}}_{\mathbf{0}}\mathbf{\right)}}$

Here,${\mathbf{a}}_{\mathbf{0}}$represents Bohr's radius and rrepresents radial distance.

Now, apply the radial kinetic energy operator on $R_{2,1}\left(r\right)$ as follows.

$-\frac{h^2}{2m}\frac{1}{r^2}\frac{d}{dr}\left(r\frac{d}{dr}\right)R_{2,1}\left(r\right)=-\frac{h^2}{2m}\left(\frac{1}{r^2}-\frac{1}{r{a}_0}+\frac{1}{4a_0^2}\right)R_{2,1}\left(r\right)$

This is because

$\frac{d}{dr}r{e}^{-\left(2a_0\right)}=e^{-r\left(2a_0\right)}\left(1+\frac{r}{-2a_0}\right)$

Thus,

$\frac{d}{dr}\left(r^2{e}^{-r\left(2a_0\right)}\left(1+\frac{r}{-2a_0}\right)\right)=e^{-r\left(2a_0\right)}\left(\left(2r+\frac{r^2}{-2a_0}\right)\left(1+\frac{r}{-2a_0}\right)+\frac{r^2}{-2a_0}\right)$

Now the expectation value for radial kinetic energy is calculated as follows.

$$\begin{gathered} {\int }_0^{\infty }-\frac{h^2}{2m}\left(\frac{2}{r^2}-\frac{2}{r{a}_0}+\frac{1}{4a_0^2}\right)R_{2,1}^2\left(r\right)dr=\left(-\frac{h^2}{2m}\frac{1}{{\left(2a_0\right)}^3}\frac{1}{3a_0^2}{\int }_0^{\infty }\left(\frac{2}{r^2}-\frac{2}{r{a}_0}+\frac{1}{4a_0^2}\right)r^2{e}^{-r/a_0}dr\right) \\ =-\frac{h^2}{2m}\frac{1}{2a_0^4}{\int }_0^{\infty }\left(2-2x+\frac{1}{4}{x}^2\right)e^{-x}dx \\ =\frac{h^2}{48m}\frac{1}{2a_0^4}\left(2-2+\frac{2}{4}\right) \\ =\frac{h^2}{96m}\frac{1}{a_0^4} \end{gathered}$$

And, the expectation value for rotational kinetic energy is as follows.

role="math" localid="1659322811595" $$\begin{gathered} {\int }_0^{\infty }-\frac{h^2}{2m}\frac{l\left(l+1\right)}{r^2}{R}_{2,1}^2\left(r\right)dr=\frac{h^2}{2m}\frac{1}{{\left(2a_0\right)}^3}\frac{1}{3a_0^2}l\left(l+1\right){\int }_0^{\infty }1e^{-r/a_0}dr \\ =\frac{h^2}{48m}\frac{1}{a_0^4}l\left(l+1\right) \end{gathered}$$

The ratio of radial kinetic energy and rotational kinetic energy is calculated as follows.

$\frac{-{\displaystyle \frac{h^2}{96m}}{\displaystyle \frac{1}{d_0^4}}}{\frac{h^2}{48m}{\displaystyle \frac{1}{a_0^{4}l\left(l+1\right)}}}=\frac{1}{2l\left(l+1\right)}$

The ratio goes to 0 as l goes to infinity.

Therefore, the ratio of expectation values of radial energy to the rotational energy is $-\frac{1}{2l\left(l+1\right)}$.