Probability of electron location in Hydrogen atom

AI Thread Summary
The discussion focuses on calculating the probability of finding an electron in the ground state of a hydrogen atom within one Bohr radius of the nucleus. The relevant formula for the radial probability density is provided, and the user correctly identifies that the electron is in the s orbital with quantum numbers n = 1 and l = 0. There is confusion regarding the variable r, which is clarified to be the radius equal to the Bohr radius (a₀) for this problem. The importance of integrating the squared radial probability density from 0 to the Bohr radius to obtain the actual probability is emphasized, along with a suggestion to use WolframAlpha for verification. The discussion concludes with the user acknowledging the need to perform the integral correctly.
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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.
 
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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
 
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