91影视

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

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

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

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

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

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

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

Here, A is the proportionality constant.

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

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

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

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

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

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

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

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

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

$\stackrel{炉}{r}={鈭�}_0^{鈭�}rP\left(r\right)dr$

Substitute $A{r}^{2n}{e}^{\frac{-2r}{n{a}_0}}$for P(r) in equation $\stackrel{炉}{r}={鈭�}_0^{鈭�}rP\left(r\right)dr$

$$\begin{gathered} \stackrel{炉}{r}={鈭�}_0^{鈭�}rA{r}^{2n}{e}^{\frac{-2r}{n{a}_0}}dr \\ =A{鈭�}_0^{鈭�}{r}^{2n+1}{e}^{\frac{-2r}{n{a}_0}}dr \end{gathered}$$

$\stackrel{炉}{r}=A{鈭�}_0^{鈭�}{r}^{2n+1}{e}^{\frac{-2r}{n{a}_0}}dr$

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

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

Substitute$\frac{\left(2n+1\right)!}{{\left({\displaystyle \frac{2}{n{a}_0}}\right)}^{2n+2}}$for${鈭�}_0^{鈭�}{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 $\stackrel{炉}{r}=A{鈭�}_0^{鈭�}{r}^{2n+1}{e}^{\frac{-2r}{n{a}_0}}dr$for A in equation.

$\stackrel{炉}{r}=A{鈭�}_0^{鈭�}{r}^{2n+1}{e}^{\frac{-2r}{n{a}_0}}dr$and solve $\stackrel{炉}{r}$

$\stackrel{炉}{r}=\left({\left(\frac{2}{n{a}_0}\right)}^{2n+1}\frac{1}{\left(2n\right)!}\right)脳\frac{(2n+1)}{{\left(\frac{2}{n{a}_0}\right)}^{2n+2}}$

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

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.

$\stackrel{炉}{r^2}={鈭�}_0^{鈭�}{r}^{2}P\left(r\right)dr$

Substitute $A{r}^{2n}{e}^{\frac{-2r}{n{a}_0}}$for $P\left(r\right)$ in equation $\stackrel{炉}{r^2}={鈭�}_0^{鈭�}{r}^{2}P\left(r\right)dr$

$\stackrel{炉}{r^2}={鈭�}_0^{鈭�}{r}^2\left(A{r}^{2n}{e}^{\frac{-2r}{n{a}_0}}\right)dr$

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

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

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

$\stackrel{炉}{r^2}=A{鈭�}_0^{鈭�}\left(r^{2n+2}{e}^{\frac{-2r}{n{a}_0}}\right)dr$and solve for $\stackrel{炉}{r^2}$

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

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

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

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

Calculate the ratio of $\frac{螖r}{\stackrel{炉}{r}}$as follows:

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

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

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

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

For larger values of n the value of $\frac{螖r}{\stackrel{炉}{r}}$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

For larger values of n the value of $\frac{螖r}{\stackrel{炉}{r}}$ is tending to 0.