The ${\mathbf{\psi }}_{\mathbf{2}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{0}}$state –2p the state in which ${\mathbf{m}}_{\mathbf{I}}\mathbf{=}\mathbf{0}$has most of its probability density along the z-axis, and so it is often referred to as a $\mathbf{2}{\mathbf{p}}_{\mathbf{z}}$state. To allow its probability density to stick out in other ways and thus facilitate various kinds of molecular bonding with other atoms, an atomic electron may assume a wave function that is an algebraic combination of multiple wave functions open to it. One such “hybrid state” is the sum ${\mathbf{\psi }}_{\mathbf{2}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{0}}\mathbf{=}{\mathbf{\psi }}_{\mathbf{2}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{-}\mathbf{1}}$(Note: Because the Schrodinger equation is a linear differential equation, a sum of solutions with the same energy is a solution with that energy. Also, normalization constants may be ignored in the following questions.)

(a) Write this wave function and its probability density in terms of r, $\mathbf{\theta }$, and $\mathbf{\varphi }$, (Use the Euler formula to simplify your result.)

(b) In which of the following ways does this state differ from its parts (i.e., ${\mathbf{\psi }}_{\mathbf{2}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{+}\mathbf{1}}$and ${\mathbf{\psi }}_{\mathbf{2}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{-}\mathbf{1}}$) and from the 2p_z state: Energy? Radial dependence of its probability density? Angular dependence of its probability density?

(c) This state is offer is often referred to as the $\mathbf{2}{\mathbf{p}}_{\mathbf{z}}$. Why?

(d) How might we produce a $\mathbf{2}{\mathbf{p}}_{\mathbf{y}}$state?

Step by step solution

01

A concept:

As you know that the probability density can be found by the square of the absolute value of the wave function. The wave function gives us the likelihood of finding an electron at a given point in space.

02

(a) Wave function and probability density:

Wave function is define by,

$$\begin{gathered} {\psi }_{2,1+1}+{\psi }_{2,1,-1}=\left(\frac{1}{a_0^{5/2}\sqrt{24}}r{e}^{-r/2a_0}\sqrt{\frac{3}{8\mathrm{\pi }}}\sin \theta e^{+i\varphi }\right)\left(\frac{1}{a_0^{5/2}\sqrt{24}}r{e}^{-r/2a_0}\sqrt{\frac{3}{8\mathrm{\pi }}}\sin \theta e^{-i\varphi }\right) \\ =\frac{1}{4a_0^{5/2}\sqrt{\mathrm{\pi }}}r{e}^{-r/2a_0}\sin \theta \cos \varphi \end{gathered}$$

Probability Density is define by,

$\left({\psi }_{2,1,+1}+{\psi }_{2,1,-1}\right)*\left({\psi }_{2,1,+1}+{\psi }_{2,1,-1}\right)=\frac{1}{16a_0^5\mathrm{\pi }}r{e}^{-r/a_0}{\sin }^2\theta {\cos }^2\varphi$

03

(b) Difference and dependence:

Energy of all n=2 states is same; hence, it does not differ in energy.

All 2p states will have the same R(r), and thus the same radial probability dependence. The wave function ${\mathrm{\psi }}_{2,1,0}$depends on $\cos \theta$,${\mathrm{\psi }}_{2,1,+1}$depends on $\sin \theta e^{+i\varphi }$,${\mathrm{\psi }}_{2,1-1}$depends on ${\mathrm{sin\theta e}}^{-\mathrm{i\varphi }}$and the wave function depends on $\mathrm{sin\theta }\hspace{0.33em}\mathrm{cos\varphi }$. Since these all differ, hence, their angular probabilities will also differ.

04

(c) Reason to be referred as 2pz:

While$\cos \theta$, the angular factor in the $2{\mathrm{p}}_{\mathrm{z}}$is large along z, the angular factor here, $\sin \theta \hspace{0.33em}\cos \varphi$, is large along x.

05

(d) Production of 2py state:

You have,

$\left({\mathrm{\psi }}_{2,1,+1}-{\mathrm{\psi }}_{2,1,-1}\right)\mathrm{\alpha }\left({\mathrm{re}}^{-\mathrm{r}/2\mathrm{a}}{\mathrm{sin\theta e}}^{+\mathrm{i\varphi }}\right)-\left({\mathrm{re}}^{-\mathrm{r}/2{\mathrm{a}}_0}{\mathrm{sin\theta e}}^{-\mathrm{i\varphi }}\right)={\mathrm{re}}^{-\mathrm{r}/2{\mathrm{a}}_0}\mathrm{sin\theta }2\mathrm{isin\varphi }$

This is large along the y-axis.

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