Probable distance in the 2p state (H atom)

In summary, it seems that the most likely distance from the origin for an electron in the 2p state of hydrogen is 4a0.
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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.

[tex]R_{21} = \frac{1}{\sqrt{24a_{0}^{3}}}\frac{r}{a_{0}}e^{-\frac{r}{2a_{0}}}[/tex]

2. Show explicitly by integration that the spherical harmonic [tex]Y_{1,1} = -\sqrt{\frac{3}{8\pi}}sin{\theta}e^{i\phi}[/tex] 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 [tex]|R_{21}^2|.r^2 = \frac{r^4}{24a_{0}^5}.e^{-\frac{r}{a_{0}}}[/tex], and find the maxima. I find then that differentiating with respect to r and setting to 0 gives solutions [tex]r = 0, r = 4a_{0}[/tex]. The maxima is then [tex]r = 4a_{0}[/tex]. 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 [tex]e^{i\phi}[/tex] happily disappears on taking the square modulus, and we are left (are we not?) with the integral

[tex]\int^{\pi}_{0}d\theta . \int^{2\pi}_{0}d\phi . sin^{3}(\theta) \frac{3}{8\pi}[/tex]

A simple trig identity dissolves the [tex]sin^3[/tex] 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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  • #2
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
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 :-)
 

1. What is the probable distance in the 2p state of a hydrogen atom?

The probable distance in the 2p state of a hydrogen atom is approximately 0.053 nanometers.

2. How is the probable distance in the 2p state of a hydrogen atom calculated?

The probable distance in the 2p state is calculated using the Bohr model, which takes into account the energy levels of the electron and the attractive force between the electron and nucleus.

3. Does the probable distance in the 2p state vary for different atoms?

Yes, the probable distance in the 2p state can vary for different atoms based on their atomic number and electron configuration.

4. What is the significance of knowing the probable distance in the 2p state of a hydrogen atom?

Knowing the probable distance in the 2p state of a hydrogen atom can help us understand the electron's behavior and energy levels in an atom. It also allows us to make predictions and calculations about chemical reactions and bonding.

5. Can the probable distance in the 2p state of a hydrogen atom be directly measured?

No, the probable distance in the 2p state cannot be directly measured. It is a theoretical value based on mathematical calculations and the Bohr model.

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