Consider the differential cross-section for elastic scattering given in eqn \(14.46 .\) At a given energy, sketch its dependence on the scattering angle \(\theta\) when the \(l=1\) partial wave dominates the scattering. Do the same for the \(l=0\) and \(l=2\) partial waves.

Short Answer

Expert verified
To summarize: for the \(l=0\) case, the differential cross-section for elastic scattering is angle-independent and hence results in a horizontal line when plotted versus the scattering angle. For \(l=1\), the dependence is linear, yielding a slanted upward line in the plot. And for \(l=2\), the dependence is quadratic, generating a parabolic line when plotted versus angle.

Step by step solution

01

Understanding the Role of \(l\)

The quantity \(l\) denotes the angular momentum quantum number. It tells us about the shape of the orbital in which the particle resides. An \(l\) value of 0 corresponds to an s-wave (spherical wave), 1 to a p-wave (potential wave), and 2 to a d-wave (differential wave). For different values of \(l\), the cross-section behavior changes.
02

Plot for \(l=0\)

For \(l=0\) or s-wave scattering, the cross-section is angle-independent. It doesn't vary with the scattering angle. This means on a graph of cross-section (on the y-axis) against scattering angle (on the x-axis), one would expect to see a horizontal line.
03

Plot for \(l=1\)

For \(l=1\) or p-wave scattering, the cross-section is linear with respect to the scattering angle. This means on a graph of cross-section (on the y-axis) against scattering angle (on the x-axis), one would expect to see a line that slants upward as the scattering angle increases.
04

Plot for \(l=2\)

For \(l=2\) or d-wave scattering, the cross-section varies quadratically with the scattering angle. This means on a graph of cross-section (on the y-axis) against scattering angle (on the x-axis), one would expect to see a parabolic line that increases with the square of the scattering angle.

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