The wave function of a state of the hydrogen atom is given by,

$\psi=\psi_{200}+2 \psi_{211}+3 \psi_{210}+\sqrt{2} \psi_{21-1}$

To normalized this function,

$<\psi | \psi>=1$

$\left\langle A\left[\psi=\psi_{200}+2 \psi_{211}+3 \psi_{210}+\sqrt{2} \psi_{21-1}\right] A[\psi\right.$

$\begin{aligned} &\left.\left.=\psi_{200}+2 \psi_{211}+3 \psi_{210}+\sqrt{2} \psi_{21-1}\right]\right\rangle=1\\ &\therefore|A|^{2}+4|A|^{2}+9|A|^{2}+2|A|^{2}=1 \end{aligned}$

$\begin{aligned} &\therefore 16|A|^{2}=1\\ &\therefore|A|^{2}=\frac{1}{16} \end{aligned}$

$\therefore A=\frac{1}{4}$

Therefore,

$\psi=\frac{1}{4} \psi_{200}+\frac{2}{4} \psi_{211}+\frac{3}{4} \psi_{210}+\frac{\sqrt{2}}{4} \psi_{21-1}$

The eigenvalue of $L_{z}$ is given as,

$L_{z} \psi_{n l m}=m \hbar \psi_{n l m}$

The expectation value of $L_{z}$ is derived as,

$\left\langle L_{z}\right\rangle=\left\langle\psi\left|L_{z}\right| \psi\right\rangle$

$\begin{aligned} &=\left(\frac{1}{4}\right)^{2}(0 . \hbar)+\left(\frac{2}{4}\right)^{2}(1 . \hbar)+\left(\frac{3}{4}\right)^{2}(0 . \hbar)+\left(\frac{\sqrt{2}}{4}\right)^{2}((-1) . \hbar)\\ &\begin{array}{l} \therefore\left\langle\psi\left|L_{z}\right| \psi\right\rangle=\frac{\hbar}{4}-\frac{\hbar}{8} \\ \therefore\left\langle\psi\left|L_{z}\right| \psi\right\rangle=\frac{\hbar}{8} \end{array} \end{aligned}$

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Wave Function

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