Question

Work on all problems. Copy and paste the excel into the answer.

4. (20 pts) Expected values versus most likely values. The radial part of the wave function for the hydrogen atom in the 2p state is given by Ran(r) = Are 206 A is a constant and agis the Bohr radius. (a) Normalize this wave function. (b) Using the computer program of your choice or excel, graph this function. (Paste a copy in your HW solutions.) Based on your graph, what is the most likely radius for the location of the electron. (c) Now calculate the most likely radius based using calculus. Show your work. (d) Calculate the average value of r of an electron in this state.

Answer 1

(a)$R_{2p}(r)=Are^{-\frac{r}{2a_B}}$

Normalising the wavefunction

$\int_0^\infty r^2 R^2_{2p}(r)dr=1$

$A^2\int_0^\infty r^4 e^{-\frac{r}{a_B}}dr=1$

Let, $\frac{r}{a_B}=x$ ,

Jo

$A=\sqrt \frac{1}{4!a_B^5}$

Hence normalised wavefunction is

$R_{2p}(r)= \sqrt \frac{1}{4!a_B^5}re^{-\frac{r}{2a_B}}$

(b)

The graph of the function is:

te

The maxima of the function will lie at $r=2a_B$ . So, it the most likely radius is$r=2a_B$

(c) $R_{2p}(r)= \sqrt \frac{1}{4!a_B^5}re^{-\frac{r}{2a_B}}$

For most likely radius,

$\frac{\mathrm{d} R_{2p}}{\mathrm{d} r}=0$

$\frac{\mathrm{d} }{\mathrm{d} r} \left (\sqrt \frac{1}{4!a_B^5}re^{-\frac{r}{2a_B}} \right )=0$

$\sqrt \frac{1}{4!a_B^5} \frac{\mathrm{d} }{\mathrm{d} r} \left (re^{-\frac{r}{2a_B}} \right )=0$

$\frac{\mathrm{d} }{\mathrm{d} r} \left (re^{-\frac{r}{2a_B}} \right )=0$

$e^{-\frac{r}{2a_B}}-\frac{r}{2a_B}e^{-\frac{r}{2a_B}}=0$

$\left (1-\frac{r}{2a_B} \right )e^{-\frac{r}{2a_B}}=0$

$\left (1-\frac{r}{2a_B} \right ) =0$

Hence, $r=2a_B$ is the most likely radius.

(d)

Average radius is given by:

$<r>=\int_0^\infty r*r^2 R^2_{2p}(r)dr$

$<r>=\int_0^\infty r^5 \frac{1}{4!a_B^5}e^{-\frac{r}{a_B}}dr$

$<r>= \frac{1}{4!a_B^5}\int_0^\infty r^5 e^{-\frac{r}{a_B}}dr$

Let, $\frac{r}{a_B}=x$ ,

$<r>= \frac{a_B^6}{4!a_B^5}\int_0^\infty x^5 e^{-x}dx$

$<r>= \frac{a_B^6}{4!a_B^5}5!$