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The ${\mathbf{蠄}}_{\mathbf{2}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{0}}$state 鈥�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 鈥渉ybrid state鈥� is the sum ${\mathbf{蠄}}_{\mathbf{2}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{0}}\mathbf{=}{\mathbf{蠄}}_{\mathbf{2}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{-}\mathbf{1}}$(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, $\mathbf{胃}$, and $\mathbf{蠒}$, (Use the Euler formula to simplify your result.)

(b) In which of the following ways does this state differ from its parts (i.e., ${\mathbf{蠄}}_{\mathbf{2}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{+}\mathbf{1}}$and ${\mathbf{蠄}}_{\mathbf{2}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{-}\mathbf{1}}$) 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,

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

Probability Density is define by,

$\left({蠄}_{2,1,+1}+{蠄}_{2,1,-1}\right)*\left({蠄}_{2,1,+1}+{蠄}_{2,1,-1}\right)=\frac{1}{16a_0^5\mathrm{蟺}}r{e}^{-r/a_0}{\sin }^2胃{\cos }^2蠒$

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 ${\mathrm{蠄}}_{2,1,0}$depends on $\cos 胃$,${\mathrm{蠄}}_{2,1,+1}$depends on $\sin 胃e^{+i蠒}$,${\mathrm{蠄}}_{2,1-1}$depends on ${\mathrm{蝉颈苍胃e}}^{-\mathrm{颈蠒}}$and the wave function depends on $\mathrm{蝉颈苍胃}鈥�\mathrm{肠辞蝉蠒}$. Since these all differ, hence, their angular probabilities will also differ.

04

(c) Reason to be referred as 2pz:

While$\cos 胃$, the angular factor in the $2{\mathrm{p}}_{\mathrm{z}}$is large along z, the angular factor here, $\sin 胃鈥�\cos 蠒$, is large along x.

05

(d) Production of 2py state:

You have,

$\left({\mathrm{蠄}}_{2,1,+1}-{\mathrm{蠄}}_{2,1,-1}\right)\mathrm{伪}\left({\mathrm{re}}^{-\mathrm{r}/2\mathrm{a}}{\mathrm{蝉颈苍胃e}}^{+\mathrm{颈蠒}}\right)-\left({\mathrm{re}}^{-\mathrm{r}/2{\mathrm{a}}_0}{\mathrm{蝉颈苍胃e}}^{-\mathrm{颈蠒}}\right)={\mathrm{re}}^{-\mathrm{r}/2{\mathrm{a}}_0}\mathrm{蝉颈苍胃}2\mathrm{颈蝉颈苍蠒}$

This is large along the y-axis.

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