Chapter 8, Problem 30P

To determine

The uncertainty product $\Delta r\Delta p$ for the $1s$ electron of a hydrogen-like atom.

Answer to Problem 30P

The uncertainty product $\Delta r\Delta p$ for the $1s$ electron of a hydrogen-like atom is $0.866\hslash$.

Explanation of Solution

To compute uncertainty in distance, first calculate average distance and average square distance using radial probability distance.

Write the expression for probability density.

    $P_{1s}\left(r\right)=\frac{4Z}{a_0^3}{r}^2{e}^{-2Zr/a_0}$

Here, $P\left(r\right)$ is the probability density, $Z$ is the atomic number, $a_0$ is the Bohr’s radius and $r$ is the radius.

Write the expression for average distance.

    $\langle r\rangle ={\displaystyle \underset{0}{\overset{\infty }{\int }}r{P}_{1s}\left(r\right)dr}$

Here, $\langle r\rangle$ is the average distance.

Substitute $\frac{4Z}{a_0^3}{r}^2{e}^{-2Zr/a_0}$ for $P_{1s}\left(r\right)$ in above equation.

  $\langle r\rangle =\frac{4Z}{a_0^3}{\displaystyle \underset{0}{\overset{\infty }{\int }}{r}^3{e}^{-2Zr/a_0}dr}$        (I)

Write the expression for Average Square of distance.

    $\langle r^2\rangle ={\displaystyle \underset{0}{\overset{\infty }{\int }}{r}^2{P}_{1s}\left(r\right)dr}$

Here, $\langle r^2\rangle$ is Average Square of distance.

Substitute $\frac{4Z}{a_0^3}{r}^2{e}^{-2Zr/a_0}$ for $P_{1s}\left(r\right)$ in above equation.

    $\langle r^2\rangle =\frac{4Z}{a_0^3}{\displaystyle \underset{0}{\overset{\infty }{\int }}{r}^4{e}^{-2Zr/a_0}dr}$        (II)

Substitute $z$ for $\frac{2Zr}{a_0}$ in equation (I).

    $\begin{array}{c}\langle r\rangle =4{\left(\frac{Z}{a_0}\right)}^3{\left(\frac{a_0}{2Z}\right)}^4{\displaystyle \underset{0}{\overset{\infty }{\int }}{z}^3{e}^{-z}dz}\\ =\frac{1}{4}\left(\frac{a_0}{Z}\right)3!\\ =\frac{3}{2}\left(\frac{a_0}{Z}\right)\end{array}$

Substitute $z$ for $\frac{2Zr}{a_0}$ in equation (II).

    $\begin{array}{c}\langle r^2\rangle =4{\left(\frac{Z}{a_0}\right)}^3{\left(\frac{a_0}{2Z}\right)}^5{\displaystyle \underset{0}{\overset{\infty }{\int }}{z}^4{e}^{-z}dz}\\ =\frac{4!}{8}{\left(\frac{a_0}{Z}\right)}^2\\ =3{\left(\frac{a_0}{Z}\right)}^2\end{array}$

Write the expression for uncertainty of radius.

    $\Delta r={\left(\langle r^2\rangle -{\langle r\rangle }^2\right)}^{1/2}$

Substitute $3{\left(\frac{a_0}{Z}\right)}^2$ for $\langle r^2\rangle$ and $\frac{3}{2}\left(\frac{a_0}{Z}\right)$ for $r$ in above equation.

    $\begin{array}{c}\Delta r={\left(3{\left(\frac{a_0}{Z}\right)}^2-\frac{9}{4}{\left(\frac{a_0}{Z}\right)}^2\right)}^{1/2}\\ =\frac{a_0}{Z}{\left(3-\frac{9}{4}\right)}^{1/2}\\ =0.866\frac{a_0}{Z}\end{array}$

Write the expression for the average potential energy.

    $\begin{array}{c}\langle U\rangle =-kZ{e}^2{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{1}{r}}{P}_{1s}\left(r\right)dr\\ =-4kZ{e}^2{\left(\frac{Z}{a_0}\right)}^3{\displaystyle \underset{0}{\overset{\infty }{\int }}r{e}^{-2Zr/a_0}dr}\\ =-4kZ{e}^2{\left(\frac{Z}{a_0}\right)}^3{\left(\frac{a_0}{2Z}\right)}^2\\ =-\frac{k{\left(Ze\right)}^2}{a_0}\end{array}$

Write the expression for average momentum.

    $\begin{array}{c}\langle p^2\rangle =2m_e\langle K\rangle \\ =2m_e\langle E-\langle U\rangle \rangle \end{array}$

Here, $\langle p\rangle$ is the average momentum, $m_e$ is the mass of electron, $E$ is the total energy and $\langle U\rangle$ is the average potential energy.

Substitute $-\frac{k{\left(Ze\right)}^2}{2a_0}$ for $E$ and $\frac{{\hslash }^2}{m_{e}k{e}^2}$ for $a_0$ .

    $\begin{array}{c}\langle p^2\rangle =\frac{2m_{e}k{\left(Ze\right)}^2}{2a_0}\\ ={\left(\frac{Z\hslash }{a_0}\right)}^2\end{array}$

From symmetry $\langle p\rangle =0$ and uncertainty in momentum is:

    $\begin{array}{c}\Delta p={\left(\langle p^2\rangle \right)}^{1/2}\\ =\frac{Z\hslash }{a_0}\end{array}$

So product of uncertainty in momentum and uncertainty in position is:

    $\begin{array}{c}\Delta r\Delta p=\left(0.866\frac{a_0}{Z}\right)\left(\frac{Z\hslash }{a_0}\right)\\ =0.866\hslash \end{array}$

This value $0.866\hslash$ is consistent for any value of $Z$  and with the uncertainty principle.

Conclusion:

Thus, the uncertainty product $\Delta r\Delta p$ for the $1s$ electron of a hydrogen-like atom is $0.866\hslash$.