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The location of an electron at a specific place in space (defined by its x, y, and z coordinates) and the amplitude of its wave, which corresponds to its energy, are related by a mathematical function known as a wave function,$\mathit{蠄}$.

The equation for the spatial wave function of a hydrogen atom ,

${蠄}_{n/m}=\sqrt{{\left(\frac{2}{na}\right)}^3\frac{\left(n-I-1\right)!}{2n{\left[\left(n+I\right)!\right]}^3}}{e}^{-r/na}{\left(\frac{2r}{na}\right)}^I{L}_{n-I-1}^{2I+1}\left(\frac{2r}{na}\right)Y_I^m\left(胃,蠒\right)$

HereL is an associated Laguerre polynomial andY is a spherical harmonic, and they are given as follow:

$$\begin{gathered} L_p^q\left(x\right)=c_0\underset{j=0}{\overset{p}{鈭�}}\frac{{\left(-1\right)}^j\left(p+q\right)!}{\left(p-j\right)!\left(q+j\right)!j!}{x}^j{Y}_I^m\left(胃,蠒\right) \\ ={\left[\frac{2I+1\left(I-m\right)!}{4\mathrm{蟺}\left(\mathrm{I}+\mathrm{m}\right)!}\right]}^{-1/2}{e}^{imo}{P}_I^m\left(\cos \left(胃\right)\right) \end{gathered}$$

Here$P_I^m$is the associated Legendre function:

$P_l^m\left(x\right)=(1-x^2{)}^{\left|m\right|/2}{\left(\frac{d}{dx}\right)}^{\left|m\right|}{P}_l(x)$

And,

$P_l\left(x\right)=\frac{1}{2^{l}l!}{\left(\frac{d}{dx}\right)}^I(x^2-1{)}^l$

For n=4,I= 3 and m =3, determine $Y_3^3$. To find it construct $Y_I^I$

$Y_l^l={(-1)}^I\sqrt{\frac{(2\mathrm{l}+1)}{4\mathrm{蟺}}\frac{1}{\left(2I\right)!}}{e}^{i}l蠒P_l^l\left(cos\right(胃\left)\right)$

From (1), write$P_I^I$as:

$P_l^l\left(x\right)=(1-x^2{)}^{1/2}{\left(\frac{d}{dx}\right)}^I{P}_l(x)$

Replace from (5) with $P_I$

$P_l^l\left(x\right)=\frac{1}{2^{l}l!}(1-x^2{)}^{I/2}{\left(\frac{d}{dx}\right)}^{2I}(x^2-1{)}^l$

but $(x^2-1{)}^l=x^{2I}+鈰�$, and the remaining term has a power less than $2I$.So, when differentiate $(x^2-1{)}^l,2l$times all the terms vanishes except the first term with the power of , thus:

$P_l^l\left(x\right)=\frac{1}{2^{l}l!}(1-x^2{)}^{I/2}{\left(\frac{d}{dx}\right)}^2{x}^{2I}$

Now,

${\left(\frac{d}{dx}\right)}^n{x}^n=n!$

Hence:

$P_l^l=\frac{\left(2l\right)!}{2^{l}l!}(1-x^2{)}^{I/2}$

Next for $x=cos\left(胃\right),1-x^2={\sin }^2\left(胃\right)$:

$P_l^l=\frac{\left(2l\right)!}{2^{l}l!}si{n}^l\left(胃\right)$

So:

$$\begin{gathered} Y_I^I={\left(-1\right)}^I\sqrt{\frac{\left(2I+1\right)}{4\mathrm{蟺}\left(2\mathrm{I}\right)!}}{e}^{iI蠒}\frac{\left(2I\right)!}{2^{I}I!}{\sin }^i\left(胃\right) \\ ={\left(-1\right)}^I\sqrt{\frac{\left(2I\right)!\left(2I+1\right)}{4\mathrm{蟺}}}{e}^{iI蠒}\frac{1}{2^{I}I!}{\sin }^I\left(胃\right) \\ =\frac{1}{I!}\sqrt{\frac{\left(2I+1\right)!}{4\mathrm{蟺}}}{\left(-\frac{1}{2}{e}^{i蠒}\sin 胃\right)}^I \end{gathered}$$

Again, forI=3,

$y_3^3=-{\left(\frac{35}{64\mathrm{蟺}}\right)}^{\frac{1}{2}}{\sin }^3\left(胃\right)e^{3i蠒}$

Also, for $n=4,l=3$and $m=3$, use $L_0^7\left(x\right)=7!=5040.$Substitute $L_0^7\left(x\right)$and$Y_3^3$into the overall formula (1),

localid="1658397501310" $$\begin{gathered} {蠄}_{433}=\sqrt{{\left(\frac{1}{2a}\right)}^3\frac{1}{6脳{5040}^3}}{e}^{-r/4a}{\left(\frac{r}{2a}\right)}^3\left(5040\right)\left(-{\left(\frac{35}{64\mathrm{蟺}}\right)}^{\frac{1}{2}}{\sin }^3\left(胃\right)e^{3i蠒}\right) \\ =-\frac{1}{6144\sqrt{{\mathrm{蟺补}}^9}}{r}^3{e}^{-r/4a}{\sin }^3胃e^{3i蠒}{蠄}_{433} \end{gathered}$$

Therefore, the wave function for hydrogen in the given states as a function of the spherical coordinates$r,胃$and$=-\frac{1}{6144\sqrt{{\mathrm{蟺补}}^9}}{r}^3{e}^{-r/4a}{\sin }^3胃e^{3i蠒}{蠄}_{433}$is .

Evaluate the expectation value of$r$, that is $\left(r\right)$, as:

localid="1658403974750" $$\begin{gathered} \left(r\right)=鈭�r{\left|蠄\right|}^2{d}^{3}r \\ =\frac{1}{{\left(6144\right)}^2{\mathrm{蟺补}}^9}鈭�r\left(r^6{e}^{-r/2a}{\sin }^6\left(胃\right)\right)r^2\sin \left(胃\right)drd胃d蠒 \\ =\frac{1}{{\left(6144\right)}^2{\mathrm{蟺补}}^9}{鈭�}_0^{鈭�}{r}^9{e}^{-r/2a}{鈭�}_0^{\mathrm{蟺}}\left(胃\right)d胃{鈭�}_0^{2\mathrm{蟺}}d蠒 \\ =\frac{1}{{\left(6144\right)}^2{\mathrm{蟺补}}^9}\left[9!{\left(2a\right)}^{10}\right]\left(2\frac{1.4.6}{3.5.7}\right)\left(2\mathrm{蟺}\right) \end{gathered}$$

Further evaluate and get,

$$\begin{gathered} =18a\left(r\right) \\ =18a \end{gathered}$$

Thus, the expectation value of r in the state is 18a.

Assume thatbe an eigen function of the operator with eigen value of $l(l+1){鈩�}^2,\mathrm{for}I=3$

Thus:

$$\begin{gathered} L^2=I\left(I+1\right){魔}^2 \\ =12{魔}^2 \end{gathered}$$

Supposerole="math" localid="1658398355830" ${蠄}_{433}$be an eigen function of the operator $L_z$with eigen value of $m{鈩�}_{tfor}m=3,$:

$L_z=3鈩�$

Hence:

role="math" localid="1658398913271" $$\begin{gathered} L_x^2+L_y^2=L^2-L_z^2 \\ =12{魔}^2-9{魔}^2 \\ =3{魔}^2 \\ =3{魔}^2 \end{gathered}$$

Thus, the required value is$3{魔}^2$