Nuclear Physics - mean-square charge radius of a uniformly charged sphere

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Homework Help Overview

The discussion revolves around the mean-square charge radius of a uniformly charged sphere, specifically focusing on the mathematical formulation and physical interpretation of the problem. The original poster seeks to understand their approach to deriving the mean-square charge radius, referencing equations related to potential energy and radial distance within the context of nuclear physics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to evaluate the mean-square radius and the correct application of formulas related to charge distributions. The original poster expresses uncertainty about their calculations and approach, while others suggest clarifying the distinction between different concepts, such as the form factor and normalization constants.

Discussion Status

Some participants have provided guidance on using statistical methods to find the normalization constant and average values. The conversation reflects a mix of interpretations and approaches, with ongoing questions about specific calculations and concepts related to the problem.

Contextual Notes

There are indications of potential confusion regarding prerequisite knowledge in quantum mechanics and statistical methods, which may affect participants' understanding of the problem. The original poster also notes their concurrent enrollment in multiple advanced physics courses, suggesting a challenging academic load.

wdednam
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Homework Statement



Show that the mean-square charge radius of a uniformly charged sphere (with radius R) is
< r^2 > = 3*R^2 / 5

Homework Equations



< r^2 > = \int \varphif* r^2 \varphii dV

\varphif* = exp(-i q dot r)

\varphii = exp(i q dot r)

where q = kf - ki
and pf = h_bar * kf
pi = h_bar * ki

r > R: V(r) = -Ze^2 / (4*pi*e0) * 1/r^2
r < R: V´(r) = -Ze^2 / (4*pi*e0*R) * (3/2 - 1/2 * (r/R)^2)

Ei = 1/2*m*vi

Er0 = J0^2/(2*m*r0^2) + V(r0) or V´(r0) according as r > R or r < R.

J0 = Ji = m*vi*b

where b is the impact parameter and is b = Ze^2/ (4*pi*e0*m*vi^2) * cot(A/2) for hyperbolic orbits, Ei > 0

The Attempt at a Solution



I basically want to know if my approach is correct. I think that I have to find < r0^2 > where r0 is the distance of closest approach to the centre of the charge distribution either inside it (where the electron feels V´(r)) or outside it (where the electron feels V(r)).

I've done some calculations using the radial energy equation, trying to solve for r0^2, but the calculations are a mess and I get a feeling I'm going about it all wrong. The problem is from Krane's Introductory Nuclear Physics, Chapter 3, problem 3.1.
 
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You are mixing concepts here, you have tried to evaluate the Form factor, F, with the potential r^2 ..

you want to evaluate the mean square radius, then use formula for distributions:
1 = A\int \rho (\vec{r})d\vec{r}

this is from mathematical statistics. Now use that the distribution of charge is constant up radius R, and for radius larger than R it is zero. Angular integration gives you 4pi, so it is easy to find the normalization constant A.

Then you recall from statistics that the average value of some quantity Q is
&lt;Q&gt; = A\int Q\rho (\vec{r})d\vec{r}

Now you have what you need.
What you did wrong was to mix this up with the Form factor F.
 
Hi Malawi Glenn,

Thank you very much, I used the equations you suggested and got the required result.

I'm concurrently registered for Stat Mech, QM, Nuclear Physics and Solid State Phys this year and realized when I attempted this problem that I should have completed Stat Mech and QM before I took Nuc Phys and SS Phys, but what's done is done and I'll just have to cope.

Thanks again.
 
well yes, QM is pre requirement for both SS phys and Nuc phys. But the math behind is just probability, continuous and discrete.
 
I know it is too late but I have a question.

How could we find the normalization constant?

We have: ρ=3/(4.pi.R^3).

I know that we should also have A=4.pi, but the integration:

1= A ∫ρ dr

does not give that result. Any ideas?
 
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