# Volume Effect of the Proton in Hydrogen Atom

1. Apr 23, 2010

### gispiamp

This Prob is from Shankar, 17.2.3
"we assumed that the proton is a point charge e. If the proton is a uniformly dense charge distribution of radius R, the interaction is modified as
V(r)= -2(e)^2/(2R) + (er)^2/(2(R)^3) r<R
= -e^2/r r>R

Calculate 1st Order shift in the ground-state energy of H, due to this modification
Assume Exp[-R/a0]~1. (Correct answer is E(1)=2(eR)^2/(5(a0)^3))"

I try ro make perturbated Hamiltonian term to use |nlm>
H=T+V=T-e^2/r+e^2/r+V=H0+H

H is V+e^2/r (this H gives zero when r>R)

and calculate 1st order purtubation. but it doesn't give correct answer

I think the method I used is something wrong.(because calculation has no error)

2. Apr 23, 2010

### nickjer

Well I am unable to get the same shift as your answer. I get:

$$\frac{16}{15}\frac{e^2R^2}{a_0^3}$$

So I have no idea where they get the 2/5 coefficient. I could be doing something wrong as well.

3. Apr 24, 2010

### gispiamp

how could you calculate it? please explain it to me

4. Apr 24, 2010

### nickjer

Well integrate your H' from 0 to R in spherical coords. Then make the approximation for your exponentials at the end.

5. Apr 25, 2010

### gispiamp

Hm.. using Hydrogen atom's e.ft, |100>, and calculating <100|H|100> in spherical coords. I tried that way before I ask. but It doesn't gives same answer.

main Integration is
$$I_n= \int {r^n}{e^(\frac{r}{a_0})}$$

and n=4,2,1. with some coeff. It doesn't give R^2 and a_0^3

Is that wrong?

may be I tried 6~10 times... I'm tired..................................

6. Apr 25, 2010

### nickjer

Well you need to write out the whole integral. And carefully solve for it. Then expand out the exponentials with the 'R' term in it. I expanded them out to first order in R, to get a similar answer they got.

7. Apr 28, 2010

### gispiamp

to get higer order of R, I expand exponential term to 1st order

but they doesn't give correct order. I_4's coefficient is R^(-3) and Integral gives R^(5) but this order vanished and R^(3) and R^(2) remains. that's the problem

let C=2R/a_0, x=2r/a_0,

(dummy r zero to n)
I_n gives -exp[-C]*$$\sum[nPr*C^(n-r)]$$+n!

and I_1+I_2+I_4 with some coeff is the wrong answer I solved.
Hm.............. I don't know what's wrong with this...