91Ó°ÊÓ

The output of the movement of the electrons in a molecule defines the potential energy. The energy that holds the atom in the covalent bond is known as potential energy.

Consider a hydrogen atom which starts out at t=0 in the following linear combination of the stationary states n=2.l=1,m=1 and n=2,l=1,m=-1 that is expressed as follows,

$\mathrm{ψ}\left(\mathrm{r},0\right)=\frac{1}{\sqrt{2}}\left({\mathrm{ψ}}_{211}+{\mathrm{ψ}}_{21-1}\right)$ …(¾±)

Write the required expressions from problem 4.11.

$$\begin{gathered} {\mathrm{ψ}}_{211}=-\frac{1}{8\sqrt{\mathrm{Ï€}}}\frac{\mathrm{r}}{{\mathrm{a}}^{5/2}}{\mathrm{e}}^{-\mathrm{r}/2\mathrm{a}}\sin \left(\mathrm{θ}\right){\mathrm{e}}^{\mathrm{¾\pm Ï•}} \\ {\mathrm{ψ}}_{21-1}=-\frac{1}{8\sqrt{\mathrm{Ï€}}}\frac{\mathrm{r}}{{\mathrm{a}}^{5/2}}{\mathrm{e}}^{-\mathrm{r}/2\mathrm{a}}\sin \left(\mathrm{θ}\right){\mathrm{e}}^{-\mathrm{¾\pm Ï•}} \end{gathered}$$

Construct$\mathrm{ψ}\left(\mathrm{r},\mathrm{t}\right)$ that is wavefunction at t, multiply equation (i) by ${\mathrm{e}}^{-{\mathrm{iE}}_2\mathrm{t}/\sout{\mathrm{h}}}$.

$\mathrm{ψ}\left(\mathrm{r},\mathrm{t}\right)=\frac{1}{\sqrt{2}}\left({\mathrm{ψ}}_{211}+{\mathrm{ψ}}_{21-1}\right){\mathrm{e}}^{{\mathrm{iE}}_2\mathrm{t}/\sout{\mathrm{h}}}$

Write the expression for the energy of both data-custom-editor="chemistry" ${\mathrm{ψ}}_{211}$and data-custom-editor="chemistry" ${\mathrm{ψ}}_{21-1}$which is same.

$\begin{array}{rcl}{\mathrm{E}}_2& =& \frac{{\mathrm{E}}_1}{{\mathrm{n}}^2}\\ & =& \frac{{\mathrm{E}}_1}{4}\\ & =& -\frac{{\sout{\mathrm{h}}}^2}{8{\mathrm{ma}}^2}\end{array}$

Adddata-custom-editor="chemistry" ${\mathrm{ψ}}_{211}$ and data-custom-editor="chemistry" ${\mathrm{ψ}}_{21-1}$.

data-custom-editor="chemistry" ${\mathrm{ψ}}_{211}+{\mathrm{ψ}}_{21-1}=\frac{1}{\sqrt{\mathrm{Ï€²¹}}}\frac{1}{8{\mathrm{a}}^2}{\mathrm{re}}^{-\mathrm{r}/2\mathrm{a}}\sin \left(\mathrm{θ}\right)\left({\mathrm{e}}^{\mathrm{¾\pm Ï•}}-{\mathrm{e}}^{-\mathrm{¾\pm Ï•}}\right)$

Usedata-custom-editor="chemistry" ${\mathrm{e}}^{\mathrm{¾\pm Ï•}}-{\mathrm{e}}^{-\mathrm{¾\pm Ï•}}=2\mathrm{isin}\left(\mathrm{Ï•}\right)$ in the above expression.

${\mathrm{ψ}}_{211}+{\mathrm{ψ}}_{21-1}=-\frac{\mathrm{i}}{\sqrt{\mathrm{Ï€²¹}}4{\mathrm{a}}^2}{\mathrm{re}}^{-\mathrm{r}/2\mathrm{a}}\sin \left(\mathrm{θ}\right)\sin \left(\mathrm{Ï•}\right)$

Thus, the value of data-custom-editor="chemistry" $\mathrm{ψ}\left(\mathrm{r},\mathrm{t}\right)$is $-\frac{\mathrm{i}}{\sqrt{2\mathrm{Ï€²¹}}4{\mathrm{a}}^2}{\mathrm{re}}^{-\mathrm{r}/2\mathrm{Î\pm }}\sin \left(\mathrm{θ}\right)\sin \left(\mathrm{Ï•}\right){\mathrm{e}}^{-{\mathrm{iE}}_2\mathrm{t}/\sout{\mathrm{h}}}$.

Write the expression for the expected value of the potential energy.

Write the value of the potential energy of the electron in the hydrogen atom.

$\mathrm{V}=-\frac{\mathrm{e}}{4{\mathrm{Ï€Î\mu }}_0}\frac{1}{\mathrm{r}}$

Substitute the above value in the expression of expected value of the potential energy.

data-custom-editor="chemistry"

From part (a) determine the modulus square of the wave function.

${\left|\mathrm{ψ}\right|}^2=\frac{1}{\left(2\mathrm{Ï€²¹}\right)\left(16{\mathrm{a}}^4\right)}{\mathrm{r}}^2{\mathrm{e}}^{-\mathrm{r}/\mathrm{a}}{\sin }^2\left(\mathrm{θ}\right){\sin }^2\left(\mathrm{Ï•}\right)$

Substitute the above value in the expression of expected value of the potential energy.

$\begin{array}{rcl}{\left|\mathrm{ψ}\right|}^2& =& \frac{1}{\left(2\mathrm{Ï€²¹}\right)\left(16{\mathrm{a}}^4\right)}\left(-\frac{{\mathrm{e}}^2}{4{\mathrm{Ï€Î\mu }}_0}\right)∫\left({\mathrm{r}}^2{\mathrm{e}}^{-\mathrm{r}/\mathrm{a}}{\sin }^2\left(\mathrm{θ}\right){\sin }^2\left(\mathrm{Ï•}\right)\right)\frac{1}{\mathrm{r}}{\mathrm{r}}^2\sin \left(\mathrm{θ}\right)\mathrm{»å°ù»åθ»åÏ•}\\ & =& \frac{1}{32{\mathrm{Ï€²¹}}^5}\left(-\frac{{\sout{\mathrm{h}}}^2}{{\mathrm{ma}}^2}\right){∫}_0^{∞}{\mathrm{r}}^2{\mathrm{e}}^{-\mathrm{r}/\mathrm{a}}d\mathrm{r}{∫}_0^{\mathrm{Ï€}}{\sin }^3\left(\mathrm{θ}\right)d\mathrm{θ}{∫}_0^{2\mathrm{Ï€}}{\sin }^2\left(\mathrm{Ï•}\right)d\mathrm{Ï•}\\ & =& \frac{{\sout{\mathrm{h}}}^2}{32{\mathrm{Ï€³¾²¹}}^6}\left(3!{\mathrm{a}}^4\right)\left(\frac{4}{3}\right)\left(\mathrm{Ï€}\right)\\ & =& \frac{{\sout{\mathrm{h}}}^2}{4{\mathrm{ma}}^2}\end{array}$

Simplify the above expression.

Thus, the expectation value of the potential energy is -6.8eV.