# Probable distance in the 2p state (H atom)

1. Dec 2, 2007

### T-7

[SOLVED] Probable distance in the 2p state (H atom)

1. The problem statement, all variables and given/known data

1. Determine the most likely distance from the origin for an electron in the 2p state of hydrogen.

$$R_{21} = \frac{1}{\sqrt{24a_{0}^{3}}}\frac{r}{a_{0}}e^{-\frac{r}{2a_{0}}}$$

2. Show explicitly by integration that the spherical harmonic $$Y_{1,1} = -\sqrt{\frac{3}{8\pi}}sin{\theta}e^{i\phi}$$ is normalised. You may use a table of integrals.

3. The attempt at a solution

1. It seems to me the simplest approach is to obtain the probability density function $$|R_{21}^2|.r^2 = \frac{r^4}{24a_{0}^5}.e^{-\frac{r}{a_{0}}}$$, and find the maxima. I find then that differentiating with respect to r and setting to 0 gives solutions $$r = 0, r = 4a_{0}$$. The maxima is then $$r = 4a_{0}$$. Does that seem sensible? I can't seem to find any textbook values out there to check it against.

2. My query about the second one is that I find no need to use a table of integrals. The $$e^{i\phi}$$ happily disappears on taking the square modulus, and we are left (are we not?) with the integral

$$\int^{\pi}_{0}d\theta . \int^{2\pi}_{0}d\phi . sin^{3}(\theta) \frac{3}{8\pi}$$

A simple trig identity dissolves the $$sin^3$$ into a couple of sin functions. And the outcome is indeed 1. Why should we need to use a table of integrals...?

Cheers!

Last edited: Dec 2, 2007
2. Dec 2, 2007

### Gokul43201

Staff Emeritus
Correct on both counts.

You can look up integrals for powers of trig functions if you want to avoid the trouble of doing a little bit of trig. IMO, it'd probably be quicker in this case to just do the trig than to hunt down the integral in a handbook.

3. Dec 3, 2007

### T-7

Ok. Thank you :-)