Question

For the more circular orbits, $\mathit{\ell }\mathbf{=}\mathit{n}\mathbf{-}\mathbf{1}$and

$\mathit{P}\mathbf{\left(}\mathit{r}\mathbf{\right)}\mathbf{\propto }{\mathit{r}}^{\mathbf{2}\mathbf{n}}{\mathit{e}}^{\mathbf{-}\mathbf{2}\mathbf{r}\mathbf{/}\mathbf{n}{\mathbf{a}}_{\mathbf{0}}}$

a) Show that the coefficient that normalizes this probability is

localid="1660047077408" ${\left(\frac{2}{n{a}_0}\right)}^{\mathbf{2}\mathbf{n}\mathbf{+}\mathbf{1}}\frac{\mathbf{1}}{\mathbf{\left(}\mathbf{2}\mathbf{n}\mathbf{\right)}\mathbf{!}}$

b) Show that the expectation value of the radius is given by

$\overline{\mathbf{r}}\mathbf{=}\mathit{n}\left(n+\frac{1}{2}\right){\mathit{a}}_{\mathbf{0}}$

and the uncertainty by

$\mathit{\Delta }\mathit{r}\mathbf{=}\mathit{n}{\mathit{a}}_{\mathbf{0}}\sqrt{\frac{\mathbf{n}}{\mathbf{2}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{4}}}$

c) What happens to the ratio$\mathit{\Delta }\mathit{r}\mathbf{/}\overline{\mathbf{r}}$in the limit of large n? Is this large-n limit what would be expected classically?

Short answer

a) The coefficient of the wave function is ${\left(\frac{2}{n{a}_0}\right)}^{2n+1}\frac{1}{\left(2n\right)!}$

b) The uncertainty in the radius is$n{a}_0\sqrt{\frac{n}{2}+\frac{1}{4}}$

c) For larger values of n the value of $\frac{\Delta r}{\overline{r}}$ is tending to 0 and the uncertainty acts classically.

Step-by-step solution

 Given data

$P\left(r\right)\propto r^{2n}{e}^{-2r/n{a}_0}$

normalization condition

The condition for the normalization of the wave function P(r) is expressed as follows:

${\int }_{\mathbf{0}}^{\infty }\mathbf{P}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{dr}\mathbf{=}\mathbf{1}$

Where, $\mathbf{P}\left(\mathbf{r}\right)$ is the probability of finding electron in orbit of radius r.

To determine the normalization constant is 2na02n+11(2n)! for probability P(r)∝r2ne-2r/na0 a)

From the given data, the wave function is as follows.

$P\left(r\right)\propto r^{2n}{e}^{-2r/n{a}_0}$

$P\left(r\right)=A{r}^{2n}{e}^{\frac{-2r}{n{a}_0}}$

Here, A is the proportionality constant.

Substitute $A{r}^{2n}{e}^{\frac{-2r}{n{a}_0}}$for $P\left(r\right)$in equation ${\int }_0^{\infty }P\left(r\right)dr=1$

$$\begin{gathered} {\int }_0^{\infty }A{r}^{2n}{e}^{\frac{-2r}{n{a}_0}}dr=1 \\ A{\int }_0^{\infty }{r}^{2n}{e}^{\frac{-2r}{n{a}_0}}dr=1 \end{gathered}$$

From the integral identities, the value of the integral ${\int }_0^{\infty }A{r}^{2n}{e}^{\frac{-2r}{n{a}_0}}dr$is as follows.

${\int }_0^{\infty }{r}^{2n}{e}^{\frac{-2r}{n{a}_0}}dr=\frac{\left(2n\right)!}{{\left(\frac{2}{n{a}_0}\right)}^{2n+1}}$

Substitute$\frac{\left(2n\right)!}{{\left(\frac{2}{n{a}_0}\right)}^{2n+1}}$for ${\int }_0^{\infty }{r}^{2n}{e}^{\frac{-2r}{n{a}_0}}dr$ in equation $A{\int }_0^{\infty }{r}^{2n}{e}^{\frac{-2r}{n{a}_0}}dr=1$ and solve for A.

$A\frac{\left(2n\right)!}{{\left(\frac{2}{n{a}_0}\right)}^{2n+1}}=1$

$A={\left(\frac{2}{n{a}_0}\right)}^{2n+1}\frac{1}{\left(2n\right)!}$

Step 4: Conclusion

Therefore, the coefficient of the wave function is ${\left(\frac{2}{n{a}_0}\right)}^{2n+1}\frac{1}{\left(2n\right)!}$

To determine the expectation radius  r¯=n(n+12)a0b)

The expression for the expectation value of radius is as follows.

$\overline{r}={\int }_0^{\infty }rP\left(r\right)dr$

Substitute $A{r}^{2n}{e}^{\frac{-2r}{n{a}_0}}$for P(r) in equation $\overline{r}={\int }_0^{\infty }rP\left(r\right)dr$

$$\begin{gathered} \overline{r}={\int }_0^{\infty }rA{r}^{2n}{e}^{\frac{-2r}{n{a}_0}}dr \\ =A{\int }_0^{\infty }{r}^{2n+1}{e}^{\frac{-2r}{n{a}_0}}dr \end{gathered}$$

$\overline{r}=A{\int }_0^{\infty }{r}^{2n+1}{e}^{\frac{-2r}{n{a}_0}}dr$

From the integral identities, the value of the integral ${\int }_0^{\infty }{r}^{2n+1}{e}^{\frac{-2r}{n{a}_0}}dr$ is as follows.

${\int }_0^{\infty }{r}^{2n+1}{e}^{\frac{-2r}{n{a}_0}}dr=\frac{(2n+1)!}{{\left(\frac{2}{n{a}_0}\right)}^{2n+2}}$

Substitute$\frac{\left(2n+1\right)!}{{\left({\displaystyle \frac{2}{n{a}_0}}\right)}^{2n+2}}$for${\int }_0^{\infty }{r}^{2n+1}{e}^{\frac{-2r}{n{a}_0}}dr$${\left(\frac{2}{n{a}_0}\right)}^{2n+1}\frac{1}{\left(2n\right)!}$and $\overline{r}=A{\int }_0^{\infty }{r}^{2n+1}{e}^{\frac{-2r}{n{a}_0}}dr$for A in equation.

$\overline{r}=A{\int }_0^{\infty }{r}^{2n+1}{e}^{\frac{-2r}{n{a}_0}}dr$and solve $\overline{r}$

$\overline{r}=\left({\left(\frac{2}{n{a}_0}\right)}^{2n+1}\frac{1}{\left(2n\right)!}\right)\times \frac{(2n+1)}{{\left(\frac{2}{n{a}_0}\right)}^{2n+2}}$

$$\begin{gathered} \overline{r}=\left({\left(\frac{2}{n{a}_0}\right)}^{2n+1}\frac{1}{\left(2n\right)!}\right)\times \frac{(2n+1)\left(2n\right)!}{{\left(\frac{2}{n{a}_0}\right)}^{2n+1}\left(\frac{2}{n{a}_0}\right)} \\ =(2n+1)\left(\frac{n{a}_0}{2}\right) \\ =n\left(n+\frac{1}{2}\right)a_0 \end{gathered}$$

Therefore, the expectation value of radius is $n\left(n+\frac{1}{2}\right)a_0$

The expression for the expectation value of square of radius is as follows.

$\overline{r^2}={\int }_0^{\infty }{r}^{2}P\left(r\right)dr$

Substitute $A{r}^{2n}{e}^{\frac{-2r}{n{a}_0}}$for $P\left(r\right)$ in equation $\overline{r^2}={\int }_0^{\infty }{r}^{2}P\left(r\right)dr$

$\overline{r^2}={\int }_0^{\infty }{r}^2\left(A{r}^{2n}{e}^{\frac{-2r}{n{a}_0}}\right)dr$

From the integral identities, the value of the integral ${\int }_0^{\infty }\left(r^{2n+2}{e}^{\frac{-2r}{n{a}_0}}\right)dr$is as follows

${\int }_0^{\infty }\left(r^{2n+2}{e}^{\frac{-2r}{n{a}_0}}\right)dr=\frac{(2n+2)!}{{\left(\frac{2}{n{a}_0}\right)}^{2n+3}}$

Substitute$\frac{(2n+2)!}{{\left(\frac{2}{n{a}_0}\right)}^{2n+3}}$for ${\int }_0^{\infty }\left(r^{2n+2}{e}^{\frac{-2r}{n{a}_0}}\right)dr$and ${\left(\frac{2}{n{a}_0}\right)}^{2n+1}\frac{1}{\left(2n\right)!}$for A in equation.

$\overline{r^2}=A{\int }_0^{\infty }\left(r^{2n+2}{e}^{\frac{-2r}{n{a}_0}}\right)dr$and solve for $\overline{r^2}$

$$\begin{gathered} \overline{r^{\text{2}}}=\left({\left(\frac{2}{n{a}_0}\right)}^{2n+1}\frac{1}{\left(2n\right)!}\right)\frac{(2n+2)!}{{\left(\frac{2}{n{a}_0}\right)}^{2n+3}} \\ =\left({\left(\frac{2}{n{a}_0}\right)}^{2n+1}\frac{1}{\left(2n\right)!}\right)\frac{(2n+2)(2n+1)\left(2n\right)!}{{\left(\frac{2}{n{a}_0}\right)}^{2n+1}{\left(\frac{2}{n{a}_0}\right)}^2} \\ =(2n+2)(2n+1)1{\left(\frac{n{a}_0}{2}\right)}^2 \\ =n^2\left(n+\frac{1}{2}\right)(n+1)a_0^2 \end{gathered}$$

Substitute $n^2\left(n+\frac{1}{2}\right)(n+1)a_0^2$for $\overline{r^{\text{2}}}$and $n\left(n+\frac{1}{2}\right)a_0$for $\overline{r}$in equation $\Delta r=\sqrt{{\overline{r}}^2-{\overline{r}}^2}.$

$$\begin{gathered} \Delta r=\sqrt{\left[n^2\left(n+\frac{1}{2}\right)(n+1)a_0^2\right]-{\left[n\left(n+\frac{1}{2}\right)a_0\right]}^2} \\ =\sqrt{\left[n^2\left(n+\frac{1}{2}\right)(n+1)a_0^2\right]-\left[n^2{\left(n+\frac{1}{2}\right)}^2{a_0}^2\right]} \\ =n{a}_0\sqrt{\left(n+\frac{1}{2}\right)(n+1)-{\left(n+\frac{1}{2}\right)}^2} \\ =n{a}_0\sqrt{\frac{n}{2}+\frac{1}{4}} \end{gathered}$$

Step 6: Conclusion

Therefore, the uncertainty in the radius is $n{a}_0\sqrt{\frac{n}{2}+\frac{1}{4}}$

 To determine the ratio   in the limit of large nc)

Calculate the ratio of $\frac{\Delta r}{\overline{r}}$as follows:

Substitute $n{a}_0\sqrt{\frac{n}{2}+\frac{1}{4}}$for $\Delta r$and $n\left(n+\frac{1}{2}\right)a_0$for $\overline{r}$in $\frac{\Delta r}{\overline{r}}$and solve for$\frac{\Delta r}{\overline{r}}$

$\frac{\Delta r}{\overline{r}}=\frac{n{a}_0\sqrt{\frac{n}{2}+\frac{1}{4}}}{n\left(n+\frac{1}{2}\right)a_0}$

$=\frac{\sqrt{\frac{n}{2}+\frac{1}{4}}}{\left(n+\frac{1}{2}\right)}$

$=\frac{1}{\sqrt{2n+1}}$

For larger values of n the value of $\frac{\Delta r}{\overline{r}}$is tending to 0.

So that, the uncertainty in radius is tending to zero. That is, there is no uncertainty in radius for circular orbits

Step 8: Conclusion

For larger values of n the value of $\frac{\Delta r}{\overline{r}}$ is tending to 0.