Chapter 29, Problem 12P

(a)

To determine

To show that the given wave function is normalized.

Answer to Problem 12P

The given wave function is normalized.

Explanation of Solution

Write the condition for the normalization.

  ${\displaystyle {\int }_0^{\infty }{\left|{\psi }_{1s}\left(r\right)\right|}^{2}dV}=1$                                                                                                       (I)

Here, dV is the elemental volume of the sphere.

Write the expression for the elemental volume.

  $dV=4\pi r^{2}dr$

Substitute $\frac{1}{\sqrt{\pi a_0{}^3}}{e}^{\frac{-r}{a_0}}$ for ${\psi }_{1s}\left(r\right)$ and $4\pi r^{2}dr$ for dV in the equation (I) to check the wave function is normalized.

  $\begin{array}{c}{\displaystyle {\int }_0^{\infty }{\left|\frac{1}{\sqrt{\pi a_0{}^3}}{e}^{\frac{-r}{a_0}}\right|}^2{r}^{2}dr}=1\\ \left(\frac{4\pi }{\pi a_0{}^3}\right){\displaystyle {\int }_0^{\infty }{\left|e^{\frac{-r}{a_0}}\right|}^2{r}^{2}dr}=1\\ \left(\frac{4\pi }{\pi a_0{}^3}\right){\displaystyle {\int }_0^{\infty }{e}^{\frac{-2r}{a_0}}{r}^{2}dr}=1\\ \left(\frac{4}{a_0{}^3}\right)\left(\frac{2}{{\left(\frac{2}{a_0}\right)}^3}\right)=1\\ 1=1\end{array}$

The integral ${\displaystyle {\int }_0^{\infty }{e}^{\frac{-2r}{a_0}}{r}^{2}dr}$ has the value $\left(\frac{2}{{\left(\frac{2}{a_0}\right)}^3}\right)$ using the gamma function concept.

Thus, the left hand side and right hand side is equal and the condition for the normalization is satisfied.

Conclusion:

Therefore, the given wave function is normalized.

(b)

To determine

The probability of the electron.

Answer to Problem 12P

The probability of the electron is $0.498$.

Explanation of Solution

Write the expression to calculate the probability.

  $P={\displaystyle {\int }_{\frac{a_0}{2}}^{\frac{3a_0}{2}}{\left|{\psi }_{1s}\left(r\right)\right|}^{2}dV}$

Substitute $\frac{1}{\sqrt{\pi a_0{}^3}}{e}^{\frac{-r}{a_0}}$ for ${\psi }_{1s}\left(r\right)$ and $4\pi r^{2}dr$ for dV in the above equation to calculate P.

  $\begin{array}{c}P={\displaystyle {\int }_{\frac{a_0}{2}}^{\frac{3a_0}{2}}{\left|\frac{1}{\sqrt{\pi a_0{}^3}}{e}^{\frac{-r}{a_0}}\right|}^{2}4\pi r^{2}dr}\\ P=\frac{4\pi }{\pi a_0{}^3}{\displaystyle {\int }_{\frac{a_0}{2}}^{\frac{3a_0}{2}}{\left|e^{\frac{-r}{a_0}}\right|}^2{r}^{2}dr}\\ P=\frac{4}{a_0{}^3}{\displaystyle {\int }_{\frac{a_0}{2}}^{\frac{3a_0}{2}}{e}^{\frac{-2r}{a_0}}{r}^{2}dr}\end{array}$

Refer the table of integrals to solve the above expression.

  $\begin{array}{c}P=\frac{4}{a_0{}^3}\left(-\frac{a_0}{2}\right){\left[e^{\frac{-2r}{a_0}}\left(r^2+a_{0}r+\frac{a_0{}^2}{2}\right)\right]}_{\frac{a_0}{2}}^{\frac{3a_0}{2}}\\ =-\frac{2}{a_0{}^2}\left[e^{\frac{-2}{a_0}\left(\frac{3a_0}{2}\right)}\left({\left(\frac{3a_0}{2}\right)}^2+a_0\left(\frac{3a_0}{2}\right)+\frac{a_0{}^2}{2}\right)-e^{\frac{-2}{a_0}\left(\frac{a_0}{2}\right)}\left({\left(\frac{a_0}{2}\right)}^2+a_0\left(\frac{a_0}{2}\right)+\frac{a_0{}^2}{2}\right)\right]\\ =-\frac{2}{a_0{}^2}\left(e^{-3}\left(\frac{17a_0{}^2}{4}\right)-e^{-1}\left(\frac{5a_0{}^2}{4}\right)\right)\\ =-2\left(0.211-0.460\right)\\ =0.498\end{array}$

Conclusion:

Therefore, the probability of the electron is $0.498$.