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Chapter 39: Problem 63

Evaluate the form factor and the Coulomb-scattering differential cross section \(d \sigma / d \Omega\) for a beam of electrons scattering off a thin spherical shell of total charge \(Z e\) and radius \(a\). Could this scattering experiment distinguish between the thin-shell and solid-sphere charge distributions? Explain.

Short Answer

Expert verified
#Answer# The form factor F(k)^2 is given by the square of the magnitude of the scattering amplitude: \(F(k)^2 = \left|\frac{Ze^2}{4\pi \epsilon_0} \int \frac{e^{ik \cdot r}}{r^2} d^3r \right|^2\) After evaluating the form factor and differential cross-section for both the thin spherical shell and a solid sphere, we can compare the results to determine if the scattering experiment can distinguish between the two charge distributions. If the resulting form factors and differential cross-sections are significantly different, then the scattering experiment can distinguish between the thin shell and solid sphere charge distributions; otherwise, it cannot.

Step by step solution

01

Find the electric field due to a thin shell

Since the charge is distributed uniformly over the surface of the sphere, we can use Gauss's law to find the electric field outside the sphere. Gauss's law states that: \(\oint \vec{E} \cdot d \vec{A} = \frac{Q_{enc}}{\epsilon_0}\) Here, \(Q_{enc}\) is the enclosed charge, and \(\epsilon_0\) is the vacuum permittivity. For a sphere of radius r (r > a), the integral on the left-hand side becomes: \( \oint \vec{E} \cdot d \vec{A} = E(r) \cdot 4\pi r^2\) As the charge enclosed by the sphere of radius r is Ze, we get: \(E(r) \cdot 4\pi r^2 = \frac{Ze}{\epsilon_0}\) Now, we can solve for the electric field E(r), \(E(r) = \frac{Ze}{ 4\pi \epsilon_0 r^2}\)
02

Calculate the scattering amplitude

We can express the scattering amplitude in terms of the Fourier transform of the electric field: \(f(k) = \frac{Ze^2}{4\pi \epsilon_0} \int \frac{e^{ik \cdot r}}{r^2} d^3r\) Here, k is the momentum transfer vector, given by \(k = k_i - k_f\) where \(k_i\) is the initial momentum and \(k_f\) is the final momentum of the scattered electron. The integral is evaluated over all space.
03

Evaluate the form factor

The form factor is given by the square of the magnitude of the scattering amplitude f(k): \(F(k)^2 = \left| \frac{Ze^2}{4\pi \epsilon_0} \int \frac{e^{ik \cdot r}}{r^2} d^3r \right|^2\) We need to evaluate this integral to calculate the form factor.
04

Calculate the differential cross-section

The differential cross-section of the Coulomb scattering is given by: \(\frac{d\sigma}{d\Omega} = \left| \frac{f(k)}{k^2} \right|^2\) Here, we'll need to substitute the expression for the scattering amplitude f(k) and evaluate it.
05

Comparison with a solid sphere

After calculating the form factor and differential cross-section for both the thin spherical shell and a solid sphere, we will compare the results to determine if this scattering experiment can distinguish between the two charge distributions. If the resulting form factors and differential cross-sections are significantly different, then the scattering experiment can distinguish between the thin shell and solid sphere charge distributions. Otherwise, it cannot.

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