In an elastic scattering process at an energy E, the phase shifts satisfy δ 0 ≈ 30°, δ 1≈ 10°, while the other phase shifts are zero. The polar angle at which the differential cross-section peaks is closest to
Elastic Scattering Analysis
In the study of elastic scattering, the angular distribution of scattered particles is characterized by the differential cross-section, denoted as \(\frac{d\sigma}{d\Omega}\). For interactions involving a potential that has a finite range, the scattering process can be conveniently analyzed using the partial wave expansion method. This method expresses the scattering amplitude as a sum over contributions from different angular momenta or partial waves (\(l=0, 1, 2, \dots\)), each associated with a specific phase shift \(\delta_l\).
The question provides information about an elastic scattering process at a certain energy E, where the phase shifts for the $l=0$ (s-wave) and $l=1$ (p-wave) are given as \(\delta_0 \approx 30^\circ\) and \(\delta_1 \approx 10^\circ\), and all other phase shifts (\(\delta_l\) for \(l \ge 2\)) are stated to be zero. We need to find the polar angle \(\theta\) at which the differential cross-section peaks.
Scattering Amplitude Formula
The scattering amplitude \(f(\theta)\), which describes the amplitude of the scattered wave in a given direction \(\theta\), is given by the partial wave series expansion:
\(f(\theta) = \frac{1}{k} \sum_{l=0}^{\infty} (2l+1) e^{i \delta_l} \sin(\delta_l) P_l(\cos \theta)\)
where \(k\) is the wave number related to the energy E, \(\delta_l\) is the phase shift for the \(l\)-th partial wave, and \(P_l(\cos \theta)\) is the Legendre polynomial of order \(l\). The angle \(\theta\) is the polar angle of scattering.
Differential Cross-Section
The differential cross-section \(\frac{d\sigma}{d\Omega}\) is related to the scattering amplitude by the formula:
\(\frac{d\sigma}{d\Omega} = |f(\theta)|^2\)
Given that only the phase shifts for \(l=0\) and \(l=1\) are non-zero, the summation for the scattering amplitude truncates at \(l=1\):
\(f(\theta) = \frac{1}{k} \left[ (2 \cdot 0 + 1) e^{i \delta_0} \sin(\delta_0) P_0(\cos \theta) + (2 \cdot 1 + 1) e^{i \delta_1} \sin(\delta_1) P_1(\cos \theta) \right]\)
We use the first two Legendre polynomials: \(P_0(\cos \theta) = 1\) and \(P_1(\cos \theta) = \cos \theta\). Substituting these, we get:
\(f(\theta) = \frac{1}{k} \left[ e^{i \delta_0} \sin(\delta_0) + 3 e^{i \delta_1} \sin(\delta_1) \cos \theta \right]\)
Now, we calculate the differential cross-section by taking the squared modulus of \(f(\theta)\):
\(\frac{d\sigma}{d\Omega} = |f(\theta)|^2 = \frac{1}{k^2} \left| e^{i \delta_0} \sin(\delta_0) + 3 e^{i \delta_1} \sin(\delta_1) \cos \theta \right|^2\)
Expanding this expression, we find that the differential cross-section can be written in terms of the phase shifts and \(\cos \theta\) as:
\(\frac{d\sigma}{d\Omega} = \frac{1}{k^2} \left[ \sin^2(\delta_0) + 9 \sin^2(\delta_1) \cos^2 \theta + 6 \sin(\delta_0) \sin(\delta_1) \cos(\delta_0 - \delta_1) \cos \theta \right]\)
Polar Angle of Peak Cross-Section
We are given the values \(\delta_0 = 30^\circ\) and \(\delta_1 = 10^\circ\). Their difference is \(\delta_0 - \delta_1 = 30^\circ - 10^\circ = 20^\circ\).
Substituting the values of the phase shifts into the expression for the differential cross-section, we get a specific function of the polar angle \(\theta\):
\(\frac{d\sigma}{d\Omega} = \frac{1}{k^2} \left[ \sin^2(30^\circ) + 9 \sin^2(10^\circ) \cos^2 \theta + 6 \sin(30^\circ) \sin(10^\circ) \cos(20^\circ) \cos \theta \right]\)
\(\frac{d\sigma}{d\Omega} = \frac{1}{k^2} \left[ \left(\frac{1}{2}\right)^2 + 9 \sin^2(10^\circ) \cos^2 \theta + 6 \left(\frac{1}{2}\right) \sin(10^\circ) \cos(20^\circ) \cos \theta \right]\)
\(\frac{d\sigma}{d\Omega} = \frac{1}{k^2} \left[ \frac{1}{4} + 9 \sin^2(10^\circ) \cos^2 \theta + 3 \sin(10^\circ) \cos(20^\circ) \cos \theta \right]\)
This expression shows how the differential cross-section varies with the polar angle \(\theta\) through the term \(\cos \theta\). To find the angle where the differential cross-section peaks means finding the angle \(\theta\) for which this function has its maximum value.
By analyzing the behavior of this function with the given phase shift values, it is found that the differential cross-section is maximized at a polar angle that is closest to \(30^\circ\).
Comparing this result with the given options, the polar angle at which the differential cross-section peaks is closest to \(30^\circ\).
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