Expected r^2 for the 2s wavefunction of hydrogen atom

Click For Summary
SUMMARY

The expected value for r^2 for the 2s wavefunction of the hydrogen atom can be calculated using the radial part of the wavefunction R(r) = 1/sqrt(2a^3)*(1-r/2a)*e^(-r/2a), where a represents the Bohr radius. The integral to solve is ∫(R(r))^2 * r^2 dr from 0 to infinity. Utilizing the gamma function, Γ(n) = ∫_0^∞ t^{n-1}e^{-t} dt = (n-1)!, can simplify the integration process. Integration by parts or differentiation under the integral sign can also be employed to evaluate the integral effectively.

PREREQUISITES
  • Understanding of quantum mechanics and wavefunctions
  • Familiarity with the hydrogen atom model
  • Knowledge of integration techniques, particularly for improper integrals
  • Basic understanding of the gamma function and its properties
NEXT STEPS
  • Study the properties of the gamma function in detail
  • Learn advanced integration techniques, including integration by parts
  • Explore graphical methods for evaluating integrals
  • Investigate the implications of wavefunctions in quantum mechanics
USEFUL FOR

Students and educators in quantum mechanics, physicists working with atomic models, and anyone interested in advanced mathematical techniques for solving integrals in physics.

chall
Messages
1
Reaction score
0

Homework Statement


Calculate the expected value for r^2 for the 2s wavefunction of the hydrogen atom (only the radial part of the function is needed for l=0). If you choose to solve this problem graphically, plot or sketch the function you integrate.


Homework Equations


R(r)=1/sqrt(2a^3)*(1-r/2a)*e^(-r/2a) where a=bohr radius



The Attempt at a Solution


to calculate, i know you integrate (from 0 to inf) as follows: int((R(r))^2*r^2 dr),
but I'm having trouble solving the integral. I'm not sure how I would solve it graphically either.

Any help is appreciated. Thanks
 
Physics news on Phys.org
If you're familiar with the gamma function, you can use

\Gamma(n) = \int_0^\infty t^{n-1}e^{-t}\,dt = (n-1)!

If you need to show that result, you can integrate by parts to prove it by induction. Or you can use this trick:

\int_0^\infty t^ne^{-\alpha t}\,dt = \int_0^\infty \left(-\frac{d}{d\alpha}\right)^n e^{-\alpha t}\,dt = \left(-\frac{d}{d\alpha}\right)^n \int_0^\infty e^{-\alpha t}\,dt

Do the integral, differentiate, and then set \alpha=1.
 
Last edited:

Similar threads

  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 21 ·
Replies
21
Views
2K
Replies
5
Views
2K
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
14
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K