Question

The wave function for the orbital in the$\mathbf{2}{\mathbf{p}}_{\mathbf{z}}$2 hydrogen atom is

${\mathbf{\psi }}_{\mathbf{2}{\mathbf{p}}_{\mathbf{z}}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\sqrt{\mathbf{2}\mathbf{\pi }}}{\left(\frac{Z}{a_0}\right)}^{\mathbf{3}\mathbf{/}\mathbf{2}}\mathbf{}\mathbf{}{\mathbf{\sigma e}}^{\mathbf{-}\mathbf{\sigma }\mathbf{/}\mathbf{2}\mathbf{}}\mathbf{cos\theta }$

Short answer

$$\begin{gathered} {\mathrm{\psi }}_{{}^{2{\mathrm{p}}_{\mathrm{z}}}}^2=2.47\times {10}^{28}\mathrm{for}\mathrm{\theta } \\ =0(\mathrm{Z}\mathrm{anis}) \\ {\mathrm{\psi }}_{{}^{2{\mathrm{p}}_{\mathrm{z}}}}^2=\mathrm{zero}\mathrm{for}\mathrm{\theta } \\ ={90}^{°}(\mathrm{xy}\mathrm{plane}) \end{gathered}$$

Step-by-step solution

Write down the values provided in the question:

$$\begin{gathered} {\mathrm{a}}_0=5.29\times {10}^{-11}\mathrm{m}\mathrm{Z}=1 \\ \mathrm{\sigma }=\frac{\mathrm{Zr}}{{\mathrm{a}}_0} \\ =\frac{\mathrm{r}}{5.29\times {10}^{-11}} \\ =\frac{5.29\times {10}^{-11}}{5.29\times {10}^{-11}} \\ =1 \end{gathered}$$

Square the equation provided and substitute the values:

$$\begin{gathered} {{\mathrm{\psi }}^2}_{2{\mathrm{p}}_{\mathrm{z}}}=\frac{1}{4\sqrt{2\mathrm{\pi }}}{\left(\frac{\mathrm{Z}}{{\mathrm{a}}_0}\right)}^{3/2}{(\mathrm{\sigma })}^2{\left({\mathrm{e}}^{-\mathrm{\sigma }/2}\right)}^2{\cos }^2\mathrm{\theta } \\ =\frac{1}{16\times 2\mathrm{\pi }}\times \frac{1}{{\left(5.29\times {10}^{-11}\right)}^3}\mathrm{x}\left(1\right){\mathrm{e}}^{-1}{\cos }^2\mathrm{\theta } \\ \mathrm{For}\mathrm{\theta }=0^{°}\mathrm{i}.\mathrm{e}\mathrm{z}-\mathrm{axis} \\ {\mathrm{\psi }}_{2{\mathrm{p}}_{\mathrm{z}}}=\frac{1}{32\times 3.14}\times \frac{1}{148\times {10}^{-33}}\mathrm{x}1\times 0.3678\times {\left(1\right)}^2=\frac{0.368\times {10}^{-33}}{32\times 3.14\times 148} \\ =2.47\times {10}^{-5}\times 10 \\ \mathrm{For}\mathrm{\theta }=0^{°},{{\mathrm{\psi }}^2}_{2{\mathrm{p}}_{\mathrm{z}}}\mathrm{is}2.47\times {10}^{28}\mathrm{whereas}\mathrm{it}\mathrm{is}\mathrm{zero}\mathrm{for}\mathrm{\theta }={90}^{°} \end{gathered}$$