Chapter 8, Problem 21P

(a)

To determine

Calculate the values for ${\psi }_{2s}\left(a_0\right)$.

Answer to Problem 21P

The values for ${\psi }_{2s}\left(a_0\right)$ is $1.57\times {10}^{14}{\text{m}}^{-\frac{3}{2}}$.

Explanation of Solution

Hydrogen atom is in the $2s$ state. Take $r=a_0$

Write the expression for normalized wave function for $2s$ subshell.

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

Here, $Z$ is the atomic number, ${\psi }_{2s\left(r\right)}$ is the wave function for $2s$ subshell, $a_0$ is the Bohr radius and $r$ is the radius of the subshell.

Substitute $a_0$ for $r$ in above equation.

    ${\psi }_{2s\left(r\right)}=\frac{1}{4\sqrt{2\pi }}{\left(\frac{Z}{a_0}\right)}^{\frac{3}{2}}\left(2-Z\right)e^{-Z/2}$

Substitute $1$ for $Z$ and $0.529\times {10}^{-10}\text{m}$ for $a_0$ .

    $\begin{array}{c}{\psi }_{2s\left(r\right)}=\frac{1}{4\sqrt{2\pi }}{\left(\frac{1}{0.529\times {10}^{-10}\text{m}}\right)}^{\frac{3}{2}}\left(2-1\right)e^{-1/2}\\ =\left(9.974\times {10}^{-2}\right){\left(1.89\times {10}^{10}\right)}^{\frac{3}{2}}\left(0.606\right)\\ =\left(0.0605\right){\left(1.89\times {10}^{10}\right)}^{\frac{3}{2}}{\text{m}}^{-\frac{3}{2}}\\ =1.57\times {10}^{14}{\text{m}}^{-\frac{3}{2}}\end{array}$

Conclusion:

Thus, the values for ${\psi }_{2s}\left(a_0\right)$ is $1.57\times {10}^{14}{\text{m}}^{-\frac{3}{2}}$.

(b)

To determine

Calculate the values for ${\left|{\psi }_{2s}\left(a_0\right)\right|}^2$.

Answer to Problem 21P

The values for ${\left|{\psi }_{2s}\left(a_0\right)\right|}^2$ is $2.465\times {10}^{28}{m}^{-3}$.

Explanation of Solution

The value of square of magnitude of wave function  is calculated below by subtituing in the expression.

    ${\left|{\psi }_{2s}\left(a_0\right)\right|}^2$

Substitute $1.57\times {10}^{14}{\text{m}}^{-\frac{3}{2}}$ for ${\psi }_{2s}\left(a_0\right)$ in above expression.

    $\begin{array}{c}{\left|{\psi }_{2s}\left(a_0\right)\right|}^2={\left|1.57\times {10}^{14}{\text{m}}^{-\frac{3}{2}}\right|}^2\\ =2.465\times {10}^{28}{m}^{-3}\end{array}$

Conclusion:

Thus, the values for ${\left|{\psi }_{2s}\left(a_0\right)\right|}^2$ is $2.465\times {10}^{28}{m}^{-3}$.

(c)

To determine

Calculate the values for $P_{2s}\left(a_0\right)$.

Answer to Problem 21P

The values for $P_{2s}\left(a_0\right)$ is $8.67\times {10}^8{\text{m}}^{\text{-1}}$.

Explanation of Solution

Write the expression for $P_{2s}\left(a_0\right)$.

    $P_{2s}\left(a_0\right)=4\pi a_0^2{\left|{\psi }_{2s}\left(a_0\right)\right|}^2$                                                                        

Substitute $0.529\times {10}^{-10}\text{m}$ for $a_0$ $2.465\times {10}^{28}{m}^{-3}$ for ${\left|{\psi }_{2s}\left(a_0\right)\right|}^2$ above expression.

    $\begin{array}{c}{P}_{2s}\left(a_0\right)=4\pi {\left(0.529\times {10}^{-10}\text{m}\right)}^2\left(2.465\times {10}^{28}{m}^{-3}\right)\\ =8.67\times {10}^8{\text{m}}^{\text{-1}}\end{array}$

Conclusion:

Thus, the values for $P_{2s}\left(a_0\right)$ is $8.67\times {10}^8{\text{m}}^{\text{-1}}$.