Probability of electron location in Hydrogen atom

Click For Summary

Homework Help Overview

The discussion revolves around calculating the probability of finding an electron in the ground state of a hydrogen atom within one Bohr radius of the nucleus. The subject area includes quantum mechanics and atomic physics.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply a formula for the radial probability density but questions the interpretation of the variable r in relation to the Bohr radius. Some participants suggest that the probability can be determined through integration, noting the importance of including the integral sign in the formula.

Discussion Status

The discussion is ongoing, with participants clarifying the need for integration to find the probability. There is acknowledgment of the original poster's understanding of the formula, and some guidance has been offered regarding the integration process.

Contextual Notes

Participants are working within the constraints of homework rules, which may limit the extent of assistance provided. There is also mention of using external tools like WolframAlpha for verification, indicating a reliance on computational resources.

leroyjenkens
Messages
616
Reaction score
49

Homework Statement


What is the probability that an electron in the ground state of hydrogen is within one Bohr radius of the nucleus?

Homework Equations


P_{nl}(r) = r^{2}|R_{nl}(r)|^{2}


The Attempt at a Solution


Since it's an electron in the ground state of a hydrogen atom, that means n = 1, and that means it's in the s orbital, which means l = 0.

So using the formula provided in the book for R_{10}(r), which is \frac{2}{(a_{0})^{\frac{3}{2}}}e^{\frac{-r}{a_{0}}}

I just square that whole thing and get \frac{4e^{\frac{-2r}{a_{0}}}}{(a_{0})^{3}}

I know the value of a_{0}, but I'm not sure what r is. Is r the radius, which happens to be the same as the Bohr radius (a_{0}) for this problem?
I want to be able to calculate an actual number instead of having an answer with variables in it.

Thanks.
 
Physics news on Phys.org
The probability is the integral of the quantity you squared. The bounds the integral should be 0 to the bohr radius... that will give you the probability of the electron being within the bohr radius. Does this make sense? Check your formula for the probability... seems as if you left off the integral entirely.
 
Thanks. Yeah that makes sense. The book shows it with a dr on both sides without the integral sign. But yeah, the integral sign should be there.

Now all I need to do is figure out how to properly do that integral. That's just a matter of time.

Thanks again.
 
Remember you can check your answer (or compute it entirely if your instructor let's you) with wolframalpha
 

Similar threads

Replies
5
Views
2K
  • · Replies 21 ·
Replies
21
Views
2K
Replies
6
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
14
Views
3K
Replies
29
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 0 ·
Replies
0
Views
2K
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K