Chapter 28, Problem 51E

(a)

To determine

The expected value of $r$.

Answer to Problem 51E

The expected value of $r$ is $\frac{3a_0}{2}$.

Explanation of Solution

Write the expression for the expectation value of any physical observable $x$.

    $\langle x\rangle ={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left|\Psi \left(r\right)\right|}^{2}x{d}^{3}r}$        (I)

Here, $\Psi \left(r\right)$ is the wave function associated with the particle.

Write the expression for the wave function of the 1s electron in hydrogen atom.

    ${\Psi }_{1s}=\frac{1}{\sqrt{\pi a_0^3}}{e}^{-r/a_0}$        (II)

Here, ${\Psi }_{1s}$ is the wave function and $a_0$ is the Bohr’s radius.

Conclusion:

Substitute $\frac{1}{\sqrt{\pi a_0^3}}{e}^{-r/a_0}$ for $\Psi \left(r\right)$ and $r$ for $x$ in equation (I).

    $\langle r\rangle =\left(\frac{1}{\pi a_0^3}\right){\displaystyle \underset{0}{\overset{\infty }{\int }}r{e}^{-2r/a_0}\left(4\pi r^{2}dr\right)}$

Simplify the above equation.

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

Use the following identity to evaluate the above equation.

    ${\displaystyle \int uvdx=u{\displaystyle \int vdx-{\displaystyle \int \left[\frac{du}{dx}{\displaystyle \int vdx}\right]dx}}}$

Rearrange the equation for the expectation value of $r$.

    $\langle r\rangle =\frac{3a_0}{2}$

Thus, the expectation value of $r$ is $\frac{3a_0}{2}$.

(b)

To determine

The expectation value of $1/r$.

Answer to Problem 51E

The expectation value of $1/r$ is $\frac{1}{a_0}$.

Explanation of Solution

Write the expression for the expectation value of any physical observable $x$.

    $\langle x\rangle ={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\left|\Psi \left(r\right)\right|}^{2}x{d}^{3}r}$

Here, $\Psi \left(r\right)$ is the wave function associated with the particle.

Write the expression for the wave function of the 1s electron in hydrogen atom.

    ${\Psi }_{1s}=\frac{1}{\sqrt{\pi a_0^3}}{e}^{-r/a_0}$

Here, ${\Psi }_{1s}$ is the wave function and $a_0$ is the Bohr’s radius.

Conclusion:

Substitute $\frac{1}{\sqrt{\pi a_0^3}}{e}^{-r/a_0}$ for $\Psi \left(r\right)$ and $1/r$ for $x$ in expression (I).

    $\langle r\rangle =\left(\frac{1}{\pi a_0^3}\right){\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{1}{r}{e}^{-2r/a_0}\left(4\pi r^{2}dr\right)}$

Simplify the above equation.

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

Use the following identity to evaluate the above equation.

    ${\displaystyle \int uvdx=u{\displaystyle \int vdx-{\displaystyle \int \left[\frac{du}{dx}{\displaystyle \int vdx}\right]dx}}}$

Rearrange the equation for the expectation value of $r$.

    $\langle r\rangle =\frac{1}{a_0}$

Thus, the expectation value of $r$ is $\frac{1}{a_0}$.

(c)

To determine

The reason why $1/\langle r\rangle$ is not equal to $\langle 1/r\rangle$.

Answer to Problem 51E

Both the values of the quantities are not same.

Explanation of Solution

Physically $\langle r\rangle$ is the expectation value or the most probable value of the radius for the electron in the ground state of hydrogen. But $\langle 1/r\rangle$ is the expectation value of the inverse of the radius for the electron.

Physically both the quantities have two different interpretations.

Conclusion:

Thus, both the values of the quantities are not same.