Probable distance in the 2p state (H atom)

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

Homework Statement



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.

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!
 
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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.
 
Gokul43201 said:
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.

Ok. Thank you :-)
 
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