How Does the Yukawa Potential Modify the Ground State Energy of a Hydrogen Atom?

[foranlogan2]

could anyone help me also with this question as i am very confused,i have read my notes but i really have a problem with this stuff making anysense, any help would be appreciated thanks



question 6))
The Yukawa potential is given by V°(gamma)(r) = (q)^2/{
4(pi)(eo)0r exp(-°(gamma)r)}, where °(gamma) is a constant. This
describes a screened Coulomb potential.

a)
Sketch the radial dependence of this potential.

b)State the radial SchrÄodinger equation for this potential

c)Assume that °(gamma) is a small parameter. Use ¯rst-order perturbation theory to esti-
mate the energy shift of the ground state of the hydrogen atom when the Coulomb
potential is replaced by the Yukawa potential.
(Hint: The perturbation is w = V(gamma)- Vo

 

[Meir Achuz]

a) Just sketch that function.
b) Just add V(gamma) to the free Schrodinger equation.
c) Integrate Delta E=\integral\psi*[V(gamma)-V(Coulomb)]psi,
where psi is the ground state of hydrogen.

 

[foranlogan2]

thanks meir achuz, when i am using my hint for perturbation is V(GAMMA) MINUS V0 COULOMB.
Is the coulomb potential for charges q1q2 = q^2

 

[nrqed]

thanks meir achuz, when i am using my hint for perturbation is V(GAMMA) MINUS V0 COULOMB.
Is the coulomb potential for charges q1q2 = q^2

Yes. So what you will happen is that if you Taylor expand the exponential of the Yukaway potential, the first term in the expansion will cancel the Coulomb potential. The next term in the expansion will then be your perturbation potential.

Patrick

 

[foranlogan2]

thanks patrick i have done as said and expanded taylor series to get VGAMMA = minus (-)coulomb + q^2gamma/4(pi)eo (perturbation) so then rearange to get w expectation equals perturbation.

but w = vgamma-vcoulomb not w =vgamma + coulomb which is where i have a problem and so can't rearange to get w expectation

 

[nrqed]

thanks patrick i have done as said and expanded taylor series to get VGAMMA = minus (-)coulomb + q^2gamma/4(pi)eo (perturbation) so then rearange to get w expectation equals perturbation.

but w = vgamma-vcoulomb not w =vgamma + coulomb which is where i have a problem and so can't rearange to get w expectation

I see. I am a bit perplexed too, then. My feeling is that there is a mistake in the question in that case.
The Yukawa potential is usually used to modelt nuclear forces between nucleons but I see here that they use it for a model of the em force. My guess is that they should have written teh Yukaway potential with a factor $-q^2$ (or as $+q_1 q_2$). That's the only way for me to make sense of the question.

 

[Meir Achuz]

The pertubation potential is V'=-q^2 [exp(-gr)-1]/r.
Don't Taylor expand, because that gives the wrong spatial dependence.

 

[nrqed]

The pertubation potential is V'=-q^2 [exp(-gr)-1]/r.

Ok, so this agrees with what I was saying: that there was a sign mistake in the Yukaway potential given in the question.

Don't Taylor expand, because that gives the wrong spatial dependence.

It's true there is no need to Taylor expand after all. But if gamma is truly a small parameter (in the sense that gamma times the Bohr radius is much smaller than one), it will not matter numerically if the Taylor expansion is performed or not, the result will be the same within the approximation used. It really does not matter if the radial dependence of the first order Taylor approximation is different from the full exponential, as long as gamma is very small.