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The probability density function (PDF) is used to define the probability of a random variable falling into a distinct range of values, as opposed to assuming a single value. The function explains the probability density function of the normal distribution and how the mean and variance exist.

The expression for the probability density of the function $\mathbf{Ψ}$ is given by,

$\mathbf{P}\mathbf{(}\mathbf{r}\mathbf{)}\mathbf{=}{\left|\mathrm{Ψ}\right|}^{\mathbf{2}}\mathbf{(}\mathbf{4}{\mathbf{Ï„Ï„°ù}}^{\mathbf{2}}\mathbf{)}$

The probability density of the function ${\mathrm{Ψ}}_{210}$ is calculated as follows.

$$\begin{gathered} P_{210}\left(r\right)={\left|{\mathrm{Ψ}}_{210}\right|}^2\left(4{\mathrm{Ï€°ù}}^2\right) \\ ={\left|\left(1/4\sqrt{2\mathrm{Ï€}}\right)\left(a^{-3/2}\right)\left(r/a\right)r^{-r/2a}\cos \mathrm{θ}\right|}^2\left(4{\mathrm{Ï€°ù}}^2\right) \\ ={\left(\frac{1}{4\sqrt{2\mathrm{Ï€}}}\right)}^2{\left(\frac{r}{a^{5/2}}\right)}^2{\left(r^{-r/2a}\cos \mathrm{θ}\right)}^2{\left(4\mathrm{Ï€}r\right)}^2 \\ =\left(\frac{1}{32\mathrm{Ï€}}\right)\left(\frac{r^2}{a^5}\right)\left(r^{-2r/2a}{\cos }^2\mathrm{θ}\right)\left(4{\mathrm{Ï€°ù}}^2\right) \end{gathered}$$

Simplify further.

$P_{210}\left(r\right)=\left(\frac{r^4}{8a^5}\right)r^{-r/a}{\cos }^2\mathrm{θ}$

Therefore, the probability density of the function ${\mathrm{Ψ}}_{210}$ is $P_{210}\left(r\right)=\left(\frac{r^4}{8a^5}\right)r^{-r/a}{\cos }^2\mathrm{θ}$.

Find the square of the function ${\mathrm{Ψ}}_{21+1}$.

$$\begin{gathered} {\left|{\mathrm{Ψ}}_{21+1}\right|}^2={\left(\left(1/8\sqrt{\mathrm{Ï€}}\right)\left(a^{-3/2}\right)\left(r/a\right)r^{-r/2a}\left(\sin \mathrm{θ}\right)e^{+i\mathrm{Ï•}}\right)}^2 \\ ={\left(\left(1/8\sqrt{\mathrm{Ï€}}\right)\left(\frac{r}{a^{5/2}}\right)r^{-r/2a}\left({\sin }^2\mathrm{θ}\right)e^{+i\mathrm{Ï•}}\right)}^2 \\ =\left(\frac{1}{64\mathrm{Ï€}}\right)\left(\frac{r^2}{a^5}\right)r^{-2r/2a}\left({\sin }^2\mathrm{θ}\right)e^{+2i\mathrm{Ï•}} \\ =\left(\frac{r^2}{64{\mathrm{Ï€²¹}}^5}\right)r^{-r/a}\left({\sin }^2\mathrm{θ}\right) \end{gathered}$$

Find the square of the function ${\mathrm{Ψ}}_{21-1}$.

$$\begin{gathered} {\left|{\mathrm{Ψ}}_{21+1}\right|}^2={\left(\left(1/8\sqrt{\mathrm{Ï€}}\right)\left(a^{-3/2}\right)\left(r/a\right)r^{-r/2a}\left(\sin \mathrm{θ}\right)e^{-i\mathrm{Ï•}}\right)}^2 \\ ={\left(\left(1/8\sqrt{\mathrm{Ï€}}\right)\left(\frac{r}{a^{5/2}}\right)r^{-r/2a}\left({\sin }^2\mathrm{θ}\right)e^{-i\mathrm{Ï•}}\right)}^2 \\ =\left(\frac{1}{64\mathrm{Ï€}}\right)\left(\frac{r^2}{a^5}\right)r^{-2r/2a}\left({\sin }^2\mathrm{θ}\right)e^{-2i\mathrm{Ï•}} \\ =\left(\frac{r^2}{64{\mathrm{Ï€²¹}}^5}\right)r^{-r/a}\left({\sin }^2\mathrm{θ}\right) \end{gathered}$$

Find the probability density of the above two functions.

$$\begin{gathered} P_{21Â\pm 1}\left(r\right)={\left|{\mathrm{Ψ}}_{21Â\pm 1}\right|}^2\left(4{\mathrm{Ï€°ù}}^2\right) \\ =\left(\frac{r^2}{64{\mathrm{Ï€²¹}}^5}\right)r^{-r/a}\left({\sin }^2\mathrm{θ}\right)\left(4{\mathrm{Ï€°ù}}^2\right) \\ =\left(\frac{r^4}{16a^5}\right)r^{-r/a}\left({\sin }^2\mathrm{θ}\right) \end{gathered}$$

Hence, the probability density of the function ${\mathrm{Ψ}}_{21+1}$ is $P_{21+1}\left(r\right)=\left(\frac{r^4}{16a^5}\right)r^{-r/a}({\sin }^2\mathrm{θ})$.

For the probability density in the first case where ${\mathrm{m}}_{\mathrm{l}}=0$, the probability decreases with radial distance from the nucleus, also the probability is proportional to the factor of ${\cos }^2\mathrm{θ}$ which is the maximum along the z-axis of which the angle is $\mathrm{θ}=0$. This is consistent with the dot plot.

For the probability density in the second and the third cases where ${\mathrm{m}}_{\mathrm{l}}=Â\pm 1$, the probability decreases with radial distance from the nucleus, also the probability is proportional to the factor of ${\sin }^2\mathrm{θ}$ which is the maximum in the xy -plane of which the angle is $\mathrm{θ}={90}^{∘}$. This is also consistent with the dot plot.

Therefore, the probability P(r) is consistent with the corresponding dot plot.

Add the three probabilities as follows.

$$\begin{gathered} P\left(r\right)=P_{210}\left(r\right)+P_{21+1}\left(r\right)+P_{21-1}\left(r\right) \\ =\left(\frac{r^4}{8a^5}\right)r^{-r/a}{\cos }^2\mathrm{θ}+\left(\frac{r^4}{16a^5}\right)r^{-r/a}\left({\sin }^2\mathrm{θ}\right)+\left(\frac{r^4}{16a^5}\right)r^{-r/a}\left({\sin }^2\mathrm{θ}\right) \\ =\left(\frac{r^4}{8a^5}\right)r^{-r/a}\left({\sin }^2\mathrm{θ}+{\cos }^2\mathrm{θ}\right) \\ =\left(\frac{r^4}{8a^5}\right)r^{-r/a} \end{gathered}$$

From the above equation it is clear that the total probability density depends only on r , and it is spherically symmetric.