For the 3d state of hydrogen, at what radius is the electron probability a maximum? Compare your answer with the radius of the Bohr orbit for (n=3).

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The state $(3d)$ has $(n=3)$ and $(l=2)$ , so let's write the radial probability for this state

P_{n l}(r) d r=r^{2}\left|R_{n l}(r)\right|^{2} d r

P_{n l}(r) =r^{2}\left|R_{n l}(r)\right|^{2}

P_{32}(r) =r^{2}\left|R_{31}(r)\right|^{2}

we can use $\textbf{Table 7.1}$ to write $(R_{32})$

\begin{align*} P_{32}(r) &=r^{2}\times \left[\frac{1}{\left(a_{0}\right)^{3 / 2}} \frac{4}{81 \sqrt{30}} \frac{r^{2}}{a_{0}^{2}} e^{-r / 3 a_{0}} \right]^{2}\\ &= \left(\dfrac{1}{\left(a_{0}\right)^{3 / 2}} \dfrac{4}{81 \sqrt{30}}\right)^{2} e^{-2 r / 3 a_{0}}\left(\frac{r^{2}}{a_{0}^{2}}\right)^{2} r^{2} \end{align*}

take the term $\left(\dfrac{1}{\left(a_{0}\right)^{3 / 2}} \dfrac{4}{81 \sqrt{30}}\right)^{2}$ to be a constant $(A^{2})$ . Hence we can write the probability as follows

P_{32}(r) = A^{2}\left(\frac{r^{6}}{a_{0}^{4}}\right) e^{-2 r / 3 a_{0}}

In order to get the radius at which the electron probability is maximum, we need to find the first derivative and to solve for $\dfrac{d P}{d r}=0$

\dfrac{d P}{d r}=A^{2}\left[ \frac{6 r^{5}}{a_{0}^{4}} e^{-2 r / 3 a_{0}} -\frac{2 r^{6}}{3 a_{0}^{5}} e^{-2 r / 3 a_{0}}\right]=0

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For the 3d state of hydrogen, at what radius is the electron probability a maximum? Compare your answer with the radius of the Bohr orbit for (n=3).

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