Question

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

#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

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), \(E(r) = \frac{Ze}{ 4\pi \epsilon_0 r^2}\)

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.

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.

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.

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.

Key concepts

Electric Field

The electric field is a vector field around a charged object, representing the force exerted on other charged particles. In the context of a thin spherical shell, the electric field outside the shell can be calculated using Gauss's law.
Gauss's law relates the electric field flowing through a closed surface to the charge enclosed by that surface. For a sphere, Gauss's law can be written as:

• \( \oint \vec{E} \cdot d \vec{A} = \frac{Q_{enc}}{\epsilon_0} \)

When the radius \( r \) of the measurement point is greater than the sphere's radius \( a \), the electric field \( E(r) \) becomes:

• \( E(r) = \frac{Ze}{4\pi \epsilon_0 r^2} \)

This indicates that the electric field outside a spherical shell is similar to that from a point charge located at the center of the sphere. Importantly, if you were inside the sphere (\( r < a \)), the electric field would be zero because the net enclosed charge is zero.

Gauss's Law

Gauss's Law is a fundamental tool in electromagnetism that allows for simplified calculations of electric fields in symmetric situations, such as spheres and cylinders. By linking the total electric flux through a closed surface with the charge enclosed within that surface, it provides a straightforward methodology for calculating electric fields without needing to solve complex integrals.
To apply Gauss's Law to our problem:

• The net electric flux \( \Phi \) through a spherical surface of radius \( r \) is given by \( \Phi = E \, (4\pi r^2) \).
• This is equivalent to \( \Phi = \frac{Ze}{\epsilon_0} \) because \( Q_{enc} = Ze \) for our spherical shell.

By equating these expressions, we derive the electric field expression for regions outside the charged sphere. Understanding Gauss's Law not only aids in calculating electric fields but also plays a critical role in electromagnetic theory and various practical applications in physics.

Form Factor

The form factor is integral to understanding the scattering of electrons off charged particles. It's a function that describes how a distribution of charge affects the scattering pattern compared to a point charge. In simpler terms, it determines the intensity of scattering as a function of the momentum transfer.For our problem, the form factor \( F(k)^2 \) is derived from the scattering amplitude \( f(k) \), which is computed using the Fourier transform of the electric field. The mathematical expression is:

• \( F(k)^2 = \left| \frac{Ze^2}{4\pi \epsilon_0} \int \frac{e^{ik \cdot r}}{r^2} d^3r \right|^2 \)

This expression tells us that the charge distribution, whether distributed as a thin shell or throughout a solid sphere, affects the amplitude and pattern of scattering. By computing the form factor, we can predict differences in how a beam of electrons is scattered, potentially allowing us to distinguish different charge distributions.

Differential Cross Section

The differential cross section \( \frac{d\sigma}{d\Omega} \) provides a measure of the probability that incoming particles, such as electrons, will scatter through a particular angle. In our context, it describes how electrons scatter when they interact with the electric field of a thin spherical shell.For Coulomb scattering, the differential cross section relates to the square of the amplitude of the scattering function \( f(k) \):

• \( \frac{d\sigma}{d\Omega} = \left| \frac{f(k)}{k^2} \right|^2 \)

This equation highlights the link between the scattering amplitude and the angle at which particles scatter. By evaluating \( \frac{d\sigma}{d\Omega} \) for different electron shell models, we can ascertain whether a scattering experiment can differentiate between a thin shell and a solid sphere. Differences in these measurements could indicate distinct charge distributions, thus helping in experimental physics to map out the structure of objects at a microscopic level.