Probability of electron location in Hydrogen atom

Click For Summary
SUMMARY

The probability of finding an electron in the ground state of a hydrogen atom within one Bohr radius is calculated using the formula P_{nl}(r) = r^{2}|R_{nl}(r)|^{2}. For the ground state (n=1, l=0), the radial wave function R_{10}(r) is given by \frac{2}{(a_{0})^{\frac{3}{2}}}e^{\frac{-r}{a_{0}}}. To find the probability, one must integrate the squared wave function from 0 to the Bohr radius (a_{0}), ensuring the integral sign is included in the calculation. Tools like WolframAlpha can assist in verifying the results.

PREREQUISITES
  • Understanding of quantum mechanics and atomic structure
  • Familiarity with the Bohr model of the hydrogen atom
  • Knowledge of integral calculus
  • Experience with wave functions and probability density functions
NEXT STEPS
  • Learn how to perform integrals involving exponential functions
  • Study the derivation of the radial wave function for hydrogen atoms
  • Explore the use of WolframAlpha for complex mathematical computations
  • Investigate the implications of quantum mechanics on electron probability distributions
USEFUL FOR

Students of quantum mechanics, physicists, and anyone studying atomic structure and electron behavior in hydrogen atoms.

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
4K
  • · 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