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

In summary, the mean-square charge radius of a uniformly charged sphere with radius R is <r^2> = 3*R^2/5. This can be found by using the distribution of charge, calculating the normalization constant, and using the average value of a quantity to find the mean-square radius. It is important to have a strong understanding of probability and quantum mechanics in order to successfully solve this problem.
  • #1

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 > = [tex]\int[/tex] [tex]\varphi[/tex]f* r^2 [tex]\varphi[/tex]i dV

[tex]\varphi[/tex]f* = exp(-i q dot r)

[tex]\varphi[/tex]i = 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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  • #2
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:
[tex]1 = A\int \rho (\vec{r})d\vec{r}[/tex]

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
[tex]<Q> = A\int Q\rho (\vec{r})d\vec{r}[/tex]

Now you have what you need.
What you did wrong was to mix this up with the Form factor F.
  • #3
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.
  • #4
well yes, QM is pre requirement for both SS phys and Nuc phys. But the math behind is just probability, continuous and discrete.
  • #5
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?
  • #6

1. What is the mean-square charge radius of a uniformly charged sphere?

The mean-square charge radius of a uniformly charged sphere is a measure of the distribution of electric charge within the sphere. It is defined as the average of the squared distance of each charge from the center of the sphere.

2. How is the mean-square charge radius calculated?

The mean-square charge radius can be calculated using the formula r^2 = Q/(4πε0E), where r is the mean-square charge radius, Q is the total charge of the sphere, ε0 is the permittivity of free space, and E is the electric field strength at the surface of the sphere.

3. Why is the mean-square charge radius important in nuclear physics?

The mean-square charge radius is important in nuclear physics because it provides information about the size and structure of atomic nuclei. It can also help scientists understand the distribution of charges within a nucleus and how they contribute to its stability.

4. How does the mean-square charge radius affect the behavior of particles in a nucleus?

The mean-square charge radius affects the behavior of particles in a nucleus by influencing the strength of the electromagnetic force between them. A larger mean-square charge radius can result in a weaker force, while a smaller mean-square charge radius can lead to a stronger force.

5. Can the mean-square charge radius be measured experimentally?

Yes, the mean-square charge radius can be measured experimentally using techniques such as electron scattering or muonic atom spectroscopy. These experiments involve bombarding the nucleus with particles and measuring the scattering patterns or energy levels, which can then be used to calculate the mean-square charge radius.

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