Chapter 42, Problem 77AP

To determine

The uncertainty in measuring the radial position of electron staying in the ground state of Hydrogen.

Answer to Problem 77AP

The uncertainty in measuring the radial position of electron of Hydrogen is $\underset{\_}{0.866a_0}$.

Explanation of Solution

Write the expression for the average of the square of distance between electron and proton.

    $\langle r^2\rangle ={\displaystyle \underset{\text{all space}}{\int }{\left|\psi \right|}^2{r}^{2}dV}$                                                                                          (I)

Here, $\langle r^2\rangle$ is the average of the square of distance between electron and proton, $\psi$ is the wave function of Hydrogen in ground state, $r$ is the distance between proton and electron, and $dV$ is the volume of the space.

Use expression (I) to find the average of square of distance between electron and proton of the Hydrogen atom in $1s$ state.

    $\langle r^2\rangle ={\displaystyle \underset{\text{all space}}{\int }{\psi }_{1s}^{\ast }{r}^2{\psi }_{1s}dV}$                                                                                 (II)

Here, ${\psi }_{1s}$ is the wave function of electron in $1s$ state of Hydrogen, ${\psi }_{1s}^{\ast }$ is the complex conjugate of ${\psi }_{1s}$.

Write the expression for the wave function of Hydrogen atom in $1s$ state.

    ${\psi }_{1s}=\frac{1}{\sqrt{\pi a_0^3}}{e}^{-\frac{r}{a_0}}$                                                                                               (III)

Here, $a_0$ is the Bohr radius of Hydrogen atom.

Write the expression for complex conjugate of ${\psi }_{1s}$.

    ${\psi }_{1s}^{\ast }=\frac{1}{\sqrt{\pi a_0^3}}{e}^{-\frac{r}{a_0}}$                                                                                                (IV)

$1s$ state has a spherical symmetry.

Write the expression for the volume of the spherical $s$ orbital state.

    $V=\frac{4}{3}\pi r^3$ (V)

Here, $V$ is the volume.

Take the derivative of volume with respect to the radius of the sphere and obtain $dV$.

    $\begin{array}{c}\frac{dV}{dr}=\frac{d}{dr}\left(\frac{4}{3}\pi r^3\right)\\ =\frac{4}{3}\pi \times 3r^2\\ =4\pi r^2\\ dV=4\pi r^{2}dr\end{array}$                                                                                              (VI)

Use expression (III), (IV) and (VI) in (II).

  $\langle r^2\rangle ={\displaystyle \underset{\text{all space}}{\int }{\left(\frac{1}{\sqrt{\pi a_0^3}}{e}^{-\frac{r}{a_0}}\right)}^2{r}^2\left(\frac{1}{\sqrt{\pi a_0^3}}{e}^{-\frac{r}{a_0}}\right)4\pi r^{2}dr}$                                         (VII)

Simplify equation (VII) and evaluate the integral from zero to infinity.

  $\langle r^2\rangle =\frac{4}{a_0^3}{\displaystyle {\int }_0^{\infty }{r}^4{e}^{\frac{-2r}{a_0}}}dr$                                                                                     (VIII)

Integral in equation (VIII) is similar to the general form ${\displaystyle {\int }_0^{\infty }{x}^n{e}^{-ax}dx}$, which have the general solution as,

    ${\displaystyle {\int }_0^{\infty }{x}^n{e}^{-ax}dx}=\frac{n!}{a^{n+1}}$                                                                                             (IX)

Compare expressions (VIII) and (IX).

    $\begin{array}{l}n=4\\ a=\frac{2}{a_0}\end{array}$

Use expressions (VIII) and (IX) to find the solution of (VIII).

  $\begin{array}{c}\langle r^2\rangle =\frac{4}{a_0^3}{\displaystyle {\int }_0^{\infty }{r}^4{e}^{\frac{-2r}{a_0}}}dr\\ =\frac{4}{a_0^3}\times \frac{4!}{{\left(\frac{2}{a_0}\right)}^{4+1}}\\ =\frac{96a_0^2}{32}\\ =3a_0^2\end{array}$                                                                                          (X)

Write the expression for the root mean square uncertainty in $r$.

    $\Delta r=\sqrt{\langle r^2\rangle -{\langle r\rangle }^2}$                                                                                             (XI)

Write the expression to find $\langle r\rangle$.

    $\langle r\rangle ={\displaystyle \underset{\text{all space}}{\int }{\psi }_{1s}^{\ast }r{\psi }_{1s}dV}$                                                                                  (XII)

Use expression (III), (IV), (VI) in expression (XIII) to find $\langle r\rangle$.

    $\langle r\rangle ={\displaystyle \underset{\text{all space}}{\int }\left(\frac{1}{\sqrt{\pi a_0^3}}{e}^{-\frac{r}{a_0}}\right)r\left(\frac{1}{\sqrt{\pi a_0^3}}{e}^{-\frac{r}{a_0}}\right)4\pi r^{2}dr}$                                         (XIII)

Simplify equation (XIII) and integrate from zero to infinity.

    $\langle r\rangle =\frac{4}{a_0^3}{\displaystyle {\int }_0^{\infty }{r}^3{e}^{\frac{-2r}{a_0}}}dr$                                                                                       (XIV)

Compare expressions (XIV) and (IX).

    $\begin{array}{l}n=3\\ a=\frac{2}{a_0}\end{array}$

Use expression (IX) to solve (XIII).

    $\begin{array}{c}\langle r\rangle =\frac{4}{a_0^3}{\displaystyle {\int }_0^{\infty }{r}^3{e}^{\frac{-2r}{a_0}}}dr\\ =\frac{4}{a_0^3}\times \frac{3!}{{\left(\frac{2}{a_0}\right)}^{3+1}}\\ =\frac{3a_0}{2}\end{array}$                                                                                          (XV)

Conclusion:

Substitute $\frac{3a_0}{2}$ for $\langle r\rangle$, and $3a_0^2$ for $\langle r^2\rangle$ in equation (XI) to find $\Delta r$.

    $\begin{array}{c}\Delta r=\sqrt{\langle r^2\rangle -{\langle r\rangle }^2}\\ =\sqrt{3a_0^2-{\left(\frac{3a_0}{2}\right)}^2}\\ =0.866a_0\end{array}$

Therefore, the uncertainty in measuring the radial position of electron of Hydrogen is $\underset{\_}{0.866a_0}$.