Question

In section 10.2 , we discussed two-lobed ${\mathit{P}}_{\mathbf{x}}\mathbf{.}{\mathit{P}}_{\mathbf{y}}$and ${\mathit{P}}_{\mathbf{z}}$ states and 4 lobed hybrid $\mathit{s}{\mathit{p}}^{\mathbf{3}}$ states. Another kind of hybrid state that sticks out in just one direction is the sp , formed from a single p state and an s state. Consider an arbitrary combination of the 2s state with the $\mathbf{2}{\mathit{p}}_{\mathbf{z}}$ state. Let us represent this by$\mathit{c}\mathit{o}\mathit{s}\mathit{\tau }{\mathit{\psi }}_{\mathbf{2}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}}\mathbf{+}\mathit{s}\mathit{i}\mathit{n}\mathit{\tau }{\mathit{\psi }}_{\mathbf{2}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{0}}$(The trig factors ensure normalization in carrying out the integral , cross terms integrate to 0.leaving$\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{\tau }\mathbf{\int }{\left|{\psi }_{2,0,0}\right|}^{\mathbf{2}}\mathit{d}\mathit{v}\mathbf{+}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{\tau }\mathbf{\int }{\left|\psi 2.1.0\right|}^{\mathbf{2}}\mathit{d}\mathit{v}\mathbf{.}$ Which is 1.)

(a) Calculate the probability that an electron in such a state would be in the +z-hemisphere.(Note: Here, the cross terms so not integrate to 0 )

(b) What value of𝛕leads to the maximum probability, and what is the corresponding ratio of${\mathit{\psi }}_{\mathbf{2}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}}$ and${\mathit{\psi }}_{\mathbf{2}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}}$ ?

(C) Using a computer , make a density (Shading) plot of the probability density-density versus r and𝛉- for the𝛕-value found in part (b).

Short answer

(a) The probability for an electron in this state in the +z hemisphere is $\frac{1}{2}-\frac{3}{8}\sin 2\tau$.

The maximum of the probability is taken when $\tau =-{45}^0$,the maximum is $\frac{1}{2}+\frac{3}{8}=\frac{7}{8}$.

Step-by-step solution

Significance of the probability

Probability is a way to gauge how likely something is to happen. Many things are difficult to forecast with absolute confidence. Using it, we can only make predictions about how probable an event is to happen, or its chance of happening.

Probability of Finding arbitrary State in the +z hemisphere

The arbitrary State is Given by $\cos \tau {\psi }_{2,0,0}+\sin \tau {\psi }_{2,1,0}$. Let the Probability of finding it in +z =P

$$\begin{gathered} P={\int }_0^{\frac{\mathrm{\pi }}{2}}\sin \theta .d\theta {\int }_0^{2\mathrm{\pi }}d\varphi \left(\cos \tau {\psi }_{2,0,0}+\sin \tau {\psi }_{2,1,0}\right)\left(\cos \tau {{\psi }^{*}}_{2,0,0}+\sin \tau {{\psi }^{*}}_{2,1,0}\right) \\ {\cos }^2\tau {\int }_0^{\frac{\mathrm{\pi }}{2}}\sin \theta .d\theta {\int }_0^{2\mathrm{\pi }}d\varphi {\left|{\psi }_{2,0,0}\right|}^2+{\sin }^2\tau {\int }_0^{\frac{\mathrm{\pi }}{2}}\sin \theta .d\theta {\int }_0^{2\mathrm{\pi }}d\varphi {\left|{\psi }_{2,1,0}\right|}^2 \\ +\cos \tau \sin \tau {\int }_0^{\frac{\mathrm{\pi }}{2}}\sin \theta .d\theta {\int }_0^{2\mathrm{\pi }}d\varphi \left({\psi }_{2,0,0}{{\psi }^{*}}_{2,1,0}+{\psi }_{2,1,0}{{\psi }^{*}}_{2,0,0}\right){\int }_0^{\infty }{r}^{2}dr{R}_{2,0}\left(r\right)R_{2,3}\left(r\right) \end{gathered}$$

We will only perform the integral for the radial direction for the cross terms since otherwise it is due to normalisation.

$$\begin{gathered} {\int }_0^{\frac{\mathrm{\pi }}{2}}\sin \theta .d\theta {\int }_0^{2\mathrm{\pi }}d\varphi {\left|{\psi }_{2,0,0}\right|}^2={\int }_0^{\frac{\mathrm{\pi }}{2}}\sin \theta .d\theta {\int }_0^{2\mathrm{\pi }}d\varphi \frac{1}{4\mathrm{\pi }}=\frac{1}{2} \\ {\int }_0^{\frac{\mathrm{\pi }}{2}}\sin \theta .d\theta {\int }_0^{2\mathrm{\pi }}d\varphi \left({\psi }_{2,0,0}{{\psi }^{*}}_{2,1,0}+{\psi }_{2,1,0}{{\psi }^{*}}_{2,0,0}\right) \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\sin \theta .d\theta {\int }_0^{2\mathrm{\pi }}d\varphi 2\left(\sqrt{\frac{1}{4\mathrm{\pi }}}\right)\left(\sqrt{\frac{3}{4\mathrm{\pi }}}\cos \theta \right) \\ =\sqrt{3}{\int }_0^{\frac{\mathrm{\pi }}{2}}\sin \theta \cos \theta .d\theta \\ =\frac{\sqrt{3}}{4}{\int }_0^{\frac{\mathrm{\pi }}{2}}\sin 2\theta .d\theta \\ =\frac{\sqrt{3}}{2} \end{gathered}$$

$$\begin{gathered} {\int }_0^{\infty }{r}^{2}dr{R}_{2,0}\left(r\right)R_{2,3}\left(r\right)={\int }_0^{\infty }{r}^{2}dr\frac{1}{{\left(2a_0\right)}^{{\displaystyle \frac{3}{2}}}}2\left(1-\frac{2}{2a_0}\right)e^{\frac{-r}{2a_0}}\frac{1}{{\left(2a_0\right)}^{\frac{3}{2}}}\times \frac{1}{{\left(2a_0\right)}^{\frac{3}{2}}}\times \frac{r}{\sqrt{3}{a}_0}{e}^{\frac{-r}{2a_0}} \\ =\frac{1}{8\sqrt{3}}\int dr\left(2r^3-r^4\right)e^{-r} \\ =-\frac{\sqrt{3}}{2} \end{gathered}$$

Probability equals

$$\begin{gathered} P=\frac{1}{2}{\cos }^2\tau +\frac{1}{2}{\sin }^2\tau +\frac{\sqrt{3}}{2}\left(-\frac{\sqrt{3}}{2}\right)\cos \tau \sin \tau \\ =\frac{1}{2}-\frac{3}{4}\cos \tau \sin \tau \\ =\frac{1}{2}-\frac{3}{8}\sin 2\tau \end{gathered}$$

So, The probability for an electron in this state in the +z hemisphere is $\frac{1}{2}+\frac{3}{8}\sin 2\tau$

To Find the Maximum of the Probability.

The maximum of the probability is taken when$\tau =-{45}^0$,the maximum is$\frac{1}{2}+\frac{3}{8}=\frac{7}{8}$ .

In this Case $\cos \tau =-\sin \tau =\frac{\sqrt{2}}{2}$,thus the corresponding ratio of two wave function is$1:-1$

Calculation of Probability Density.

The Probability density as a function of $r,\theta$is equal to, the radius is in units of $a_0$…

$$\begin{gathered} P\left(r,\theta \right)=\left({\psi }_{2,0,0}+{\psi }_{2,1,0}\right)\left(\psi {\times }_{2,0,0}+\psi {\times }_{2,1,0}\right) \\ \propto {\left(2\left(1-\frac{r}{2}\right)e^{\frac{-r}{2}}-r\cos \theta e^{\frac{-r}{2}}\right)}^2 \\ ={\left(2-r-r\cos \theta \right)}^2{e}^{-r} \end{gathered}$$

The Plot is as follows,where $\theta$changes from 0 to $\mathrm{\pi }$from Top to bottom. The x-axis is theRadius in term of $a_0$.From the plot we can see that the probability density is largest when $\theta$is between 0 and $\frac{\mathrm{\pi }}{2}$,that is +z axis..