Question

Calculate the uncertainties in r for the 2s and 2p states using the formula

$\mathit{\Delta }\mathit{r}\mathbf{=}\sqrt{\overline{{\mathbf{r}}^{\mathbf{2}}}\mathbf{-}{\overline{\mathbf{r}}}^{\mathbf{2}}}$

What insight does the difference between these two uncertainties convey about the nature of the corresponding orbits?

Short answer

The uncertainty in r for the 2s state$\sqrt{6}{a}_0$

The uncertainty in r for the 2p state$\sqrt{5}{a}_0$

The uncertainty is smaller for the p – state when compared to the s – state

Step-by-step solution

 Given data

Consider 2s and 2p states

 Uncertainty

Uncertainty is the interval of possible values of a measurement within which the true value of measurement lies. The uncertainty in position $\left(\Delta r\right)$ for a measurement of position $\left(\mathbf{r}\right)$is given as-$\mathbf{\Delta r}\mathbf{=}\sqrt{\overline{{\mathbf{r}}^{\mathbf{2}}}\mathbf{-}{\overline{\mathbf{r}}}^{\mathbf{2}}}$.

 To determine the uncertainties in r for the 2p state

For 2s state: $n=2\text{and}l=0$the radial solution for the 2s state is as follows.

$R_{2s}=\frac{2}{{\left(2a_0\right)}^{\frac{3}{2}}}\left(1-\frac{r}{2a_0}\right)e^{-\frac{r}{2a_0}}$

For 2p state: $n=2\text{and}l=1$the radial solution for the 2p state is as follows.

$R_{2p}=\frac{2}{{\left(2a_0\right)}^{\frac{3}{2}}}\left(\frac{r}{\sqrt{3}{a}_0}\right)e^{-\frac{r}{2a_0}}$

Here,$a_0$is the Bohr radius and its value is$a_0=0.0529\text{nm}$

For 2s state

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

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

Here P(r) is the probability per unit radial distance and the value of P(r) is

$P\left(r\right)=r^2{R}^2\left(r\right)$

Substitute $\frac{2}{{\left(2a_0\right)}^{\frac{3}{2}}}\left(1-\frac{r}{2a_0}\right)e^{-\frac{r}{2a_0}}$for R(r) in equation$P\left(r\right)=r^2{R}^2\left(r\right)$

$$\begin{gathered} P\left(r\right)=r^2{\left(\frac{2}{{\left(2a_0\right)}^{\frac{3}{2}}}\left(1-\frac{r}{2a_0}\right)e^{-\frac{r}{2a_0}}\right)}^2 \\ =\frac{r^2}{2a_0^3}\left(1-\frac{r}{a_0}+\frac{r^2}{4a_0^2}\right)e^{-\frac{r}{a_0}} \\ =\frac{1}{2a_0^3}\left(r^2-\frac{r^3}{a_0}+\frac{r^4}{4a_0^2}\right)e^{-\frac{r}{a_0}} \end{gathered}$$

Substitute $\frac{1}{2a_0^3}\left(r^2-\frac{r^3}{a_0}+\frac{r^4}{4a_0^2}\right)e^{-\frac{r}{a_0}}$for $P\left(r\right)$in equation $\overline{r}={\int }_0^{\infty }rP\left(r\right)dr$then $\overline{r}$will be

$$\begin{gathered} \overline{r}={\int }_0^{\infty }\left(r\right)\frac{1}{2a_0^3}\left(r^2-\frac{r^3}{a_0}+\frac{r^4}{4a_0^2}\right)e^{-\frac{r}{a_0}}dr \\ =\frac{1}{2a_0^3}{\int }_0^{\infty }\left(r^3-\frac{r^4}{a_0}+\frac{r^5}{4a_0^2}\right)e^{-\frac{r}{a_0}}dr \\ =\frac{1}{2a_0^3}\left(\frac{3!}{{\left(1/a_0\right)}^4}-\frac{1}{a_0}\left(\frac{4!}{{\left(1/a_0\right)}^5}\right)+\frac{1}{4a_0}\left(\frac{5!}{{\left(1/a_0\right)}^6}\right)\right) \\ =6a_0 \end{gathered}$$

The expression for the expectation value of $r^2$is as follows.

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

Substitute $\frac{1}{2a_0^3}\left(r^2-\frac{r^3}{a_0}+\frac{r^4}{4a_0^2}\right)e^{-\frac{r}{a_0}}$for P(r) in equation $\overline{r^2}={\int }_0^{\infty }{r}^{2}P\left(r\right)dr$then $\overline{r^2}$will be

$$\begin{gathered} \overline{r^2}={\int }_0^{\infty }\left(r^2\right)\frac{1}{2a_0^3}\left(r^2-\frac{r^3}{a_0}+\frac{r^4}{4a_0^2}\right)e^{-\frac{r}{a_0}}dr \\ =\frac{1}{2a_0^3}{\int }_0^{\infty }\left(r^4-\frac{r^5}{a_0}+\frac{r^6}{4a_0^2}\right)e^{-\frac{r}{a_0}}dr \end{gathered}$$

$$\begin{gathered} \overline{r^2}=\frac{1}{2a_0^3}\left(\frac{4!}{{\left(1/a_0\right)}^{\text{5}}}-\frac{1}{a_0}\left(\frac{5!}{{\left(1/a_0\right)}^6}\right)+\frac{1}{4a_0}\left(\frac{6!}{{\left(1/a_0\right)}^7}\right)\right) \\ =42a_0^2 \end{gathered}$$

To determine the uncertainties in r for the 2s state

$5a_0$The expression for the uncertainty in r is as follows

$\Delta r=\sqrt{\overline{r^2}-{\overline{r}}^2}$

Substitute $42a_0^2$for $\overline{r^2}$and $6a_0$for $\overline{r}$in equation $\Delta r=\sqrt{\overline{r^2}-{\overline{r}}^2}$then $\Delta r$will be

$$\begin{gathered} \Delta r=\sqrt{\left(42a_0^2\right)-{\left(6a_0\right)}^2} \\ =\sqrt{\left(42a_0^2\right)-{\left(36a_0^2\right)}^2} \\ =\sqrt{6}{a}_0 \end{gathered}$$

Therefore, the uncertainty in r for 2s state$\sqrt{6}{a}_0$

For 2p state:

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

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

Here$P\left(r\right)$is the probability per unit radial distance and the value of$P\left(r\right)$is

$P\left(r\right)=r^2{R}^2\left(r\right)$

Substitute $\frac{2}{{\left(2a_0\right)}^{\frac{3}{2}}}\left(\frac{r}{\sqrt{3}{a}_0}\right)e^{-\frac{r}{2a_0}}$ for R(r) in equation $P\left(r\right)=r^2{R}^2\left(r\right)$

$$\begin{gathered} P\left(r\right)=r^2{\left(\frac{2}{{\left(2a_0\right)}^{\frac{3}{2}}}\left(\frac{r}{\sqrt{3}{a}_0}\right)e^{-\frac{r}{2a_0}}\right)}^2 \\ =\left(\frac{r^4}{24a_0^5}\right)e^{-\frac{r}{a_0}} \end{gathered}$$

$P\left(r\right)=\frac{1}{2a_0^3}\left(r^2-\frac{r^3}{a_0}+\frac{r^4}{4a_0^2}\right)e^{-\frac{r}{a_0}}$

Substitute $\left(\frac{r^4}{24a_0^5}\right)e^{-\frac{r}{a_0}}$for P(r) in equation $\overline{r}={\int }_0^{\infty }rP\left(r\right)dr$, then $\overline{r}$will be

$$\begin{gathered} \overline{r}={\int }_0^{\infty }r\left(\frac{r^4}{24a_0^5}\right)e^{-\frac{r}{a_0}}dr \\ =\frac{1}{24a_0^5}{\int }_0^{\infty }{r}^5{e}^{-\frac{r}{a_0}}dr \\ =\frac{1}{24a_0^5}\left(\frac{5!}{{\left(1/a_0\right)}^6}\right) \\ =5a_0 \end{gathered}$$

The expression for the expectation value of r is as follows

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

Substitute$\left(\frac{r^4}{24a_0^5}\right)e^{-\frac{r}{a_0}}$for P(r) in equation $\overline{r^2}={\int }_0^{\infty }{r}^{2}P\left(r\right)dr$then$\overline{r^2}$will be

$$\begin{gathered} \overline{r^2}={\int }_0^{\infty }{r}^2\left(\frac{r^4}{24a_0^5}\right)e^{-\frac{r}{a_0}}dr \\ =\frac{1}{24a_0^5}{\int }_0^{\infty }{r}^6{e}^{-\frac{r}{{\displaystyle a_0}}}dr \\ =\frac{1}{24a_0^5}\left(\frac{6!}{{\left(1/a_0\right)}^7}\right) \\ =30a_0^2 \end{gathered}$$

Substitute$30a_0^2$for $\overline{r^2}$and $5a_0$for $\overline{r}$in equation $\Delta r=\sqrt{\overline{r^2}-{\overline{r}}^2}$then$\Delta r$ will be

$$\begin{gathered} \Delta r=\sqrt{\left(30a_0^2\right)-{\left(5a_0\right)}^2} \\ =\sqrt{\left(30a_0^2\right)-\left(25a_0^2\right)} \\ =\sqrt{5}{a}_0 \end{gathered}$$

Therefore, the uncertainty in r for 2p state $\sqrt{5}{a}_0$

Step 5: Conclusion

The uncertainty is smaller for the p-state when compared to the s-state.

The s-state has only radial kinetic energy. So that, the particle is oscillating through the origin and passing through many r values.

The p-state has large angular momentum and rotational kinetic energy. So that the particle is oscillating in circular orbit within less indefinite radius.