# Why human calculation is still (sometimes) better than WolframAlpha

The last post relied heavily on WolframAlpha to calculate the average distance of the 1s electron from the hydrogen nucleus. As mentioned, this integral could be done by hand by “differentiating under the integral sign” as Feynman taught many to do and as referenced in the post the Calculus of Friendship.

When it was time to calculate the average distance of the 2s electron, the (free version) of the WolframAlpha computational knowledge engine, stalled:

So in this post, we will go back to basics and calculate this integral from scratch using Feynman’s method.

To get the average radius of the 2s electron, the relevant integral to consider is still: $R_{2s} = \int_0^{2\pi}\int_0^{\pi}\int_0^\infty{r^3\cdot{\sin{\phi}}\cdot{\psi_{2s}}^2}drd{\phi}d{\theta}$. The ${\psi_{R_{2s}}}^2$ term is now: $\left(\frac{1}{4\sqrt{2\pi}a_0^{3/2}}(2-\frac{r}{a_0})e^{-\frac{r}{2a_0}}\right)^2$. This is clear from the normalized hydrogen wavefunctions table below.

To solve this hairy integral it will be necessary to distribute the square and separate the integrands :
$R_{2s} = \int_0^{2\pi}\int_0^{\pi}\int_0^\infty{r^3\cdot{\sin{\phi}}\cdot{\left(\frac{1}{4\sqrt{2\pi}a_0^{3/2}}(2-\frac{r}{a_0})e^{-\frac{r}{2a_0}}\right)^2}}drd{\phi}d{\theta}$
$= \int_0^{2\pi}\int_0^{\pi}\int_0^\infty{r^3\cdot{\sin{\phi}}\cdot{\frac{1}{32\pi\cdot{a_0^{3}}}(4-4\frac{r}{a_0}+\frac{r^2}{a_0^2})e^{-\frac{r}{a_0}}}}drd{\phi}d{\theta}$
$= \frac{1}{32\pi{a_0^{3}}}\int_0^{2\pi}\int_0^{\pi}{\sin{\phi}}\int_0^\infty{r^3}\left(4-4\frac{r}{a_0}+\frac{r^2}{a_0^2}\right)e^{-\frac{r}{a_0}}drd{\phi}d{\theta}$
$= \frac{4\pi}{32\pi{a_0^{3}}}\int_0^\infty\left(4{r^3}-4\frac{r^4}{a_0}+\frac{r^5}{a_0^2}\right)e^{-\frac{r}{a_0}}dr$
$= \frac{1}{8{a_0^{3}}}\left(\int_0^\infty4{r^3}e^{-\frac{r}{a_0}}dr-\int_0^\infty4\frac{r^4}{a_0}e^{-\frac{r}{a_0}}dr+\int_0^\infty\frac{r^5}{a_0^2}e^{-\frac{r}{a_0}}dr\right)$

We can now apply the result $\int_0^{\infty}t^{n}e^{-at}dt={\left(\frac{1}{a}\right)}^{n+1}n!$ from the “differentiation under the integral” trick mentioned above to each integral:

$= \frac{1}{8{a_0^{3}}}\left(4\cdot{3!}a_0^4-4\cdot{4!}\frac{a_0^5}{a_0}+5!\frac{a_0^6}{a_0^2}\right)$

From here we can see that the average radius of the 2s electron, $R_{2s}$ is:

$R_{2s}= \frac{a_0}{8}\left(24-96+120\right)$

$R_{2s}= 6{a_0}$

This result confirms the expectation that the average radius of the electron is proportional to the square of the principle quantum number, n: see $r=n^2a_0$ in the figure below

since we have found that: $\frac{R_{2s}}{R_{1s}} = \frac{6a_0}{\frac{3}{2}a_0} = 4 = \frac{2^2}{1^2} = \frac{n_{2s}^2}{n_{1s}^2}$

Now that we have gotten the distance correct, how can we use this work to estimate the spectral lines produced by the emission of light from the excitation of hydrogen? Hmm…