# Electron wavefunction in hydrogen

• hicetnunc
In summary, the probability of an electron being in a particular state is given by the triple integral:$$\int_0^{r_b} \int_0^{2\pi} \int_0^\pi |\psi (r)|^2 r^2 \sin{\theta} \,d\theta \,d\phi \,dr\approx \frac{1}{\pi a_0^3}\int_0^{r_b} \int_0^{2\pi} \int_0^\pi e^{-2r/a_0} r^2 \sin{\theta} \,d\theta \,d #### hicetnunc Homework Statement For a hydrogen atom, the normalised wavefunction of an electron in the 1s state, assuming a point nucleus, is$$\psi (r) = \left( \frac {1}{\pi a_0^3} \right)^{1/2} e^{-r/a_0}, where ##a_0## is the Bohr radius. Find an approximate expression for the probability of finding the electron in a small sphere of radius ##r_b << a_0## centered on the proton. What is the electronic charge density? Relevant Equations None. Hi. I would love if someone could check my solution since me and the answer sheet I found online don't agree. The probability is given by the triple integral \begin{align*} \int_0^{r_b} \int_0^{2\pi} \int_0^\pi |\psi (r)|^2 r^2 \sin{\theta} \,d\theta \,d\phi \,dr &= \frac{1}{\pi a_0^3}\int_0^{r_b} \int_0^{2\pi} \int_0^\pi e^{-2r/a_0} r^2 \sin{\theta} \,d\theta \,d\phi \,dr\\ &= \frac{4}{a_0^3} \int_0^{r_b} r^2 e^{-2r/a_0} \,dr \\ &= \frac{4}{a_0^3} \left( -\frac{a_0}{2} e^{-2r/a_0} \left( r^2 +a_0r +\frac{a_0^2}{2} \right) \right) \Bigg|_0^{r_b} \\ &= 1 - \frac{2}{a_0^2} e^{-2r_b/a_0} \left( r_b^2 +a_0 r_b + \frac{a_0^2}{2} \right) \\ &\approx 1 - e^{-2r_b/a_0} \end{align*} As for the electron charge density, ##\rho##, I figure that since \begin{align*} \int_0^{r_b} \int_0^{2\pi} \int_0^\pi \rho r^2 \sin{\theta} \,d\theta \,d\phi \,dr &= Q \end{align*} where ##Q## is the total charge, then ##\rho## must be equal to ##e|\psi|^2##, where ##e## is the elementary charge. But according to the answer sheet I found online, the probability is given by \begin{align*} \int_0^{r_b} |\psi|^2 r^2 \,dr \approx \frac{r_b^3}{\pi a_0^3} e^{-2r_b/a_0} \sim \left( \frac{r_b}{a_0} \right)^3 \end{align*} I find the answer sheet dubious, since it only integrates over the radius and skips the angular part. In addition, I am unable to see how it solves the integral to get that solution, but I've been wrong many times before. Anyone who can provide input? Edit: Accidentally posted mid-edit, sorry! Last edited: Good so far. How does one usually approximate a function close to the origin? You should have learned this in Calculus class. Or roughly graph the integrand and figure a way to approximate it. (these two methods should lead you to the same answer) • hicetnunc hutchphd said: Good so far. How does one usually approximate a function close to the origin? You should have learned this in Calculus class. Or roughly graph the integrand and figure a way to approximate it. (these two methods should lead you to the same answer) I can get the approximation ##1-e^{-2r_b/a_0}##, which seems okay-ish to me since it at least is 0 at ##r_b=0## and approaches 1 as ##r_b## approaches infinity, even though the approximation is only supposed to be used for small ##r_b##. I only used the reasoning thatr_b^2 + a_0 r_b + \frac{a_0^2}{2} = r_b^2 + a_0\left( r_b + \frac{a_0}{2} \right) \approx \frac{a_0^2}{2},$$since ##r_b^2 \approx 0## and ##r_b+a_0/2 \approx a_0/2## if ##r_b<<a_0##. Are you thinking Taylor expansion? It's been a while since my calculus days. But the most glaring problem I have with this exercise is that I use a triple integral and the answer sheet only uses a single integral. I will inevitably always have a factor of ##4\pi## that the answer sheet doesn't have. Even if I could get the same approximation as the answer sheet, there is something it and I disagree with on a fundamental level. Thanks for looking at my problem! hicetnunc said: You need to think quick and dirty here. Just expand the integrand in$$= \frac{4}{a_0^3} \int_0^{r_b} r^2 e^{-2r/a_0} \,dr \\ to leading order in r and write down the answer. In your subsequent work you did not do the more complicated expansion uniformly and therefore got inconsistent answers which is to be expected.

• hicetnunc
I see! I thought about Taylor expanding after performing the integral, but actually expanding the integrand! Ok, Taylor expanding ##\left(\ f(x) = f(a) + f'(a)(x-a) + \frac{1}{2}f''(a)(x-a)^2 + ...\ , \text{with }a=0\right)## the integrand:

\begin{align*}
\frac{d}{dr} \left( r^2 e^{-2r/a_0} \right) &= \left( 2r - \frac{2}{a_0}r^2 \right) e^{-2r/a_0} \\
\frac{d^2}{dr^2} \left( r^2 e^{-2r/a_0} \right) &= \left( 2 - \frac{8}{a_0}r + \frac{4}{a_0^2} r^2 \right) e^{-2r/a_0}
\end{align*}

and I get ##r^2 e^{-2r/a_0} = 0 + 0 (r-0) + \frac{1}{2} \cdot 2 (r-0)^2 + ...## and I'm left with

\begin{align*}
\frac{4}{a_0^3} \int_0^{r_b} r^2 e^{-2r/a_0}\,dr \approx&\ \frac{4}{a_0^3} \int_0^{r_b} r^2 \,dr \\
&= \frac{4}{3a_0^3} r^3 |_0^{r_b} \\
&= \frac{4}{3} \left( \frac{r_b}{a_0} \right)^3
\end{align*}

which is indeed similar to ##\left( \frac{r_b}{a_0} \right)^3##. Thank you so much!

Still, I wonder about the triple integral I used when the answer sheet didn't. I essentially have a factor of ##4\pi## that the answer sheet doesn't have, even if I "approximate" it away (I guess the answer sheet only takes the radial probability, while I take both the radial and angular probability?). And regarding approximating away factors like this, I seem to recall something from another course I took that said this can be done for really large numbers, but not small numbers?

But anyway, now I see how to get the same answer. Thanks again!

Edit: an integrand was missing. Added to calculation. Whoops!

hicetnunc said:
I seem to recall something from another course I took that said this can be done for really large numbers, but not small numbers?
There are various useful approximations, some of which are for large (asymptotic) values. In my experience the Taylor expansion for small deviations is by far the most used.

hutchphd said:
There are various useful approximations, some of which are for large (asymptotic) values. In my experience the Taylor expansion for small deviations is by far the most used.
Oh, I meant like using the approximation that ##3 \cdot 10^{50} \approx 10^{50}## as valid, while an approximation of ##3 \cdot 10^{-10} \approx 10^{-10}## would be invalid (in essence, for small numbers one shouldn't remove factors like that). The Taylor expansion have indeed been the most used approximation during my studies still.

Still, any thoughts about me integrating over both the radial and angular parts, while the answer sheet only uses the radial part? It's been bugging me a little, since it seems like an important detail.

Your method as shown is correct. I shan't comment on the answer sheet which I haven't seen!

• hicetnunc