Question

A hydrogen atom starts out in the following linear combination of the stationary states n=2, l=1, m=1 and n=2, l=1, m=-1.

$\mathbf{\psi }\left(r,0\right)\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\left({\psi }_{211}+{\psi }_{21-1}\right)$

(a) Construct$\mathbf{\psi }\left(r,t\right)$Simplify it as much as you can.

(b) Find the expectation value of the potential energy,$<V>$. (Does it depend on t?) Give both the formula and the actual number, in electron volts.

Short answer

(a) The value of $\mathrm{\psi }\left(\mathrm{r},\mathrm{t}\right)$is $-\frac{1}{\sqrt{2\mathrm{\pi a}}4{\mathrm{a}}^2}{\mathrm{re}}^{-\mathrm{r}/2\mathrm{\alpha }}\sin \left(\mathrm{\theta }\right)\sin \left(\mathrm{\varphi }\right){\mathrm{e}}^{-{\mathrm{iE}}_2\mathrm{t}/\sout{\mathrm{h}}}$.

(b) The expectation value of the potential energy is -6.8eV.

Step-by-step solution

Definition of potential energy

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.

(a) Construction of ψ(r,t)

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,

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

Write the required expressions from problem 4.11.

$$\begin{gathered} {\mathrm{\psi }}_{211}=-\frac{1}{8\sqrt{\mathrm{\pi }}}\frac{\mathrm{r}}{{\mathrm{a}}^{5/2}}{\mathrm{e}}^{-\mathrm{r}/2\mathrm{a}}\sin \left(\mathrm{\theta }\right){\mathrm{e}}^{\mathrm{i\varphi }} \\ {\mathrm{\psi }}_{21-1}=-\frac{1}{8\sqrt{\mathrm{\pi }}}\frac{\mathrm{r}}{{\mathrm{a}}^{5/2}}{\mathrm{e}}^{-\mathrm{r}/2\mathrm{a}}\sin \left(\mathrm{\theta }\right){\mathrm{e}}^{-\mathrm{i\varphi }} \end{gathered}$$

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

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

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

data-custom-editor="chemistry" ${\mathrm{\psi }}_{211}+{\mathrm{\psi }}_{21-1}=\frac{1}{\sqrt{\mathrm{\pi a}}}\frac{1}{8{\mathrm{a}}^2}{\mathrm{re}}^{-\mathrm{r}/2\mathrm{a}}\sin \left(\mathrm{\theta }\right)\left({\mathrm{e}}^{\mathrm{i\varphi }}-{\mathrm{e}}^{-\mathrm{i\varphi }}\right)$

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

${\mathrm{\psi }}_{211}+{\mathrm{\psi }}_{21-1}=-\frac{\mathrm{i}}{\sqrt{\mathrm{\pi a}}4{\mathrm{a}}^2}{\mathrm{re}}^{-\mathrm{r}/2\mathrm{a}}\sin \left(\mathrm{\theta }\right)\sin \left(\mathrm{\varphi }\right)$

Thus, the value of data-custom-editor="chemistry" $\mathrm{\psi }\left(\mathrm{r},\mathrm{t}\right)$is $-\frac{\mathrm{i}}{\sqrt{2\mathrm{\pi a}}4{\mathrm{a}}^2}{\mathrm{re}}^{-\mathrm{r}/2\mathrm{\alpha }}\sin \left(\mathrm{\theta }\right)\sin \left(\mathrm{\varphi }\right){\mathrm{e}}^{-{\mathrm{iE}}_2\mathrm{t}/\sout{\mathrm{h}}}$.

(b) Determination of the expectation value of potential energy (V)

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.

$\mathrm{V}=-\frac{\mathrm{e}}{4{\mathrm{\pi \epsilon }}_0}\frac{1}{\mathrm{r}}$

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.

${\left|\mathrm{\psi }\right|}^2=\frac{1}{\left(2\mathrm{\pi a}\right)\left(16{\mathrm{a}}^4\right)}{\mathrm{r}}^2{\mathrm{e}}^{-\mathrm{r}/\mathrm{a}}{\sin }^2\left(\mathrm{\theta }\right){\sin }^2\left(\mathrm{\varphi }\right)$

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

$\begin{array}{rcl}{\left|\mathrm{\psi }\right|}^2& =& \frac{1}{\left(2\mathrm{\pi a}\right)\left(16{\mathrm{a}}^4\right)}\left(-\frac{{\mathrm{e}}^2}{4{\mathrm{\pi \epsilon }}_0}\right)\int \left({\mathrm{r}}^2{\mathrm{e}}^{-\mathrm{r}/\mathrm{a}}{\sin }^2\left(\mathrm{\theta }\right){\sin }^2\left(\mathrm{\varphi }\right)\right)\frac{1}{\mathrm{r}}{\mathrm{r}}^2\sin \left(\mathrm{\theta }\right)\mathrm{drd\theta d\varphi }\\ & =& \frac{1}{32{\mathrm{\pi a}}^5}\left(-\frac{{\sout{\mathrm{h}}}^2}{{\mathrm{ma}}^2}\right){\int }_0^{\infty }{\mathrm{r}}^2{\mathrm{e}}^{-\mathrm{r}/\mathrm{a}}d\mathrm{r}{\int }_0^{\mathrm{\pi }}{\sin }^3\left(\mathrm{\theta }\right)d\mathrm{\theta }{\int }_0^{2\mathrm{\pi }}{\sin }^2\left(\mathrm{\varphi }\right)d\mathrm{\varphi }\\ & =& \frac{{\sout{\mathrm{h}}}^2}{32{\mathrm{\pi ma}}^6}\left(3!{\mathrm{a}}^4\right)\left(\frac{4}{3}\right)\left(\mathrm{\pi }\right)\\ & =& \frac{{\sout{\mathrm{h}}}^2}{4{\mathrm{ma}}^2}\end{array}$

Simplify the above expression.

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