Distance between electron and proton in Hydrogen

Click For Summary

Homework Help Overview

The discussion revolves around calculating the probability of finding an electron more than 0.1 nm away from a proton in hydrogen atoms, specifically in the ground state described by the wave function ##\psi_{1,0,0}##. Participants explore the relationship between the calculated probability and the Bohr radius, questioning the validity of their results and the definitions involved.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of the probability using integrals and the implications of the Bohr radius as an expected value for the distance between the electron and proton. Questions arise regarding the correctness of signs in expressions and the interpretation of the Bohr radius.

Discussion Status

There is an ongoing examination of the calculations, with some participants correcting each other’s expressions. Clarifications about the Bohr radius and its significance in relation to the electron-proton distance are being sought, indicating a productive exploration of the topic.

Contextual Notes

Participants express uncertainty about the definitions and implications of the Bohr radius, indicating a need for further clarification on its role in quantum mechanics and its historical context in atomic models.

skrat
Messages
740
Reaction score
8

Homework Statement


I had to calculate the probability that we find electron more than 0.1 nm away from proton in Hydrogen atoms if ##\psi _{n,l,m}=\psi _{1,0,0}## (i don't know the english word for this state, but I think we all know what we are talking about :D)
My result is 0.09957.

Now I am questioning myself if that makes any sense at all... Correct me if I am wrong: Bohr radius is expected value for distance between electron and proton in Hydrogen atoms for ##\psi _{1,0,0}##.
If that is true, and if ##r_B=5.29*10^{-11}m## than my result makes some sense...

Homework Equations


The Attempt at a Solution



##\psi _{1,0,0}=R_{1,0}(r)Y_{0,0}(\theta ,\varphi )## where ##R_{1,0}=\frac{1}{\sqrt{4\pi }}\frac{2}{r^{3/2}}e^{-\frac{r}{r_B}}##

than ##P(r>0.1 nm)=\int_{0.1 nm}^{\infty }\frac{1}{4\pi }\frac{4}{r^{3}}e^{-\frac{2r}{r_B}}4\pi r^2dr##

##P(r>0.1 nm)=\frac{1}{2}\int_{\frac{0.2 nm}{r_B}}^{\infty }u^2e^{-u} dr##

##P(r>0.1 nm)=\frac{1}{2}e^{\frac{0.2nm}{r_B}}(4(\frac{0.1 nm}{r_B})^{2}-4(\frac{0.1 nm}{r_B})+2)##

which should be 0.09957...

Does this sound ok?
 
Physics news on Phys.org
In your last expression, are you sure all of the signs are correct in the factor multiplying the exponential?

Also, is there a typo in the sign of the argument of the exponential?
 
TSny said:
In your last expression, are you sure all of the signs are correct in the factor multiplying the exponential?

Also, is there a typo in the sign of the argument of the exponential?

true, there is a mistake, let's say that ##r_0=0.1 nm## than

##P(r>r_0)=\frac{1}{2}\int_{\frac{2r_0}{r_B}}^{\infty }u^2e^{-u} dr##

this is now, according to a book: ##\int x^2e^{ax}=e^{ax}(\frac{x^2}{a}-2\frac{x}{a^2}+2\frac{1}{a^3})## in my case a=-1

so ## P(r>r_0)=\frac{1}{2}(0-e^{-\frac{2r_0}{r_B}}(-\frac{4r_{0}^{2}}{r_{B}^{2}}-4\frac{r_{0}}{r_{B}}-2)##

which means that you are right...

##P(r>r_0)=\frac{1}{2} e^{-\frac{2r_0}{r_B}}(\frac{4r_{0}^{2}}{r_{B}^{2}}+4\frac{r_{0}}{r_{B}}+2)=0.272##

right?
 
That looks good to me.
 
Just one question:
Does bohr radius have any relation with the distance between the electron and proton?
 
I'm not sure what you are asking. Can you be more specific?
 
Is Bohr radius the average distance between the proton and electron or is it not?

In other words: what is bohr radius? I can't remember if my professor told us exactly what it is, so to me, at this moment, Bohr radius is just a number I don't really understand...
 
Historically, the bohr radius is the radius of the circular orbit of the ground state in the old "Bohr model" of the hydrogen atom.

In the quantum mechanical description, the bohr radius happens to be the distance r from the proton at which the radial probability density is maximum.
 
Thanks!
 
  • #10

Similar threads

Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
Replies
5
Views
2K
  • · Replies 21 ·
Replies
21
Views
2K
Replies
14
Views
3K
Replies
29
Views
2K
  • · Replies 5 ·
Replies
5
Views
820
  • · Replies 6 ·
Replies
6
Views
5K
Replies
6
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K