Use the Born approximation to calculate the differential cross-section for scattering from the spherical square-well potential (Section 14.5 ). Hint. Use integration by parts to determine the scattering amplitude.

Short Answer

Expert verified
By employing the Born approximation and performing an integration by parts, we can evaluate the scattering amplitude caused by a spherical square well potential. Then, using this scattering amplitude, we are able to calculate the differential cross-section for such scattering.

Step by step solution

01

Setup the Born Approximation

The Born approximation for the first order in the potential is represented by, \( f^{(1)}(k',k) = -\frac{2m}{\hbar ^2} \int V(r)e^{iqr}d^3r \), where \( q=k'-k \) is the change in momentum, \( r \) is the radius and \( V(r) \) is the potential defined by the well. The spherical coordinates are then employed to solve this integral.
02

Define the scattering amplitude

Express the scattering amplitude as \( f^{(1)}_l(k,q)= -\frac{2m}{\hbar ^2} \frac{\sqrt{4\pi}}{q} \int^\infty_0 rn(r)j_0(qr)dr \), where \( j_0(qr) \) is the spherical Bessel function of first kind. The function \( n(r) \) is the normalized radial eigenfunction of the unperturbed Hamiltonian.
03

Evaluate the integral for scattering amplitude

Use integration by parts to solve for \( f^{(1)}_l(k,q) \). Integration by parts states \( \int u dv = u(v) - \int v du \) where \( u = rn(r) \) and \( dv = j_0(qr)dr \). You can now calculate the scattering amplitude.
04

Calculate the cross-section

Once the scattering amplitude is computed, we can determine the cross-section using the formula \( \frac{d\sigma} {d\omega} = |f^{(1)}_l(k,q)|^2 \)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Equation 14.3 gives the form of the S matrix for a one-dimensional system in which a particle is scattered from an abrupt blip in the potential energy. Write down the analogous expression for scattering from a comparable dip in the potential energy. Proceed to compute the transmission probability for positive energies given that the particle is incident from the left.

Find an expression for the WKB phase shift for S-wave scattering at an energy \(E\) by the potential \(V=A / r^{2}.\)

For elastic scattering by a central potential, it is possible to show analytically that if the potential is repulsive, with \(V(r)>0\) for all \(r,\) then the scattering phase shift \(\delta_{l}(E)\) is negative; likewise, if the potential is attractive, with \(V(r)<0\) for all \(r,\) then the phase shift \(\delta_{l}\) is positive. Explain this result qualitatively by considering the effect of a repulsive (or attractive) potential on the wavelength of the scattered particle.

Show for the elastic scattering of a particle by a central potential \(V(r)\) that approaches zero more rapidly than \(1 / r\) as \(r \rightarrow \infty\) that the integral cross-section can be written as $$\sigma_{\mathrm{tot}}=\frac{4 \pi}{k} \mathrm{im} f_{k}(0)$$ where im \(f_{k}(0)\) is the imaginary part of the forward scattering amplitude \((\theta=0) .\) This is the so-called optical theorem. Hint. The Legendre polynomials are required to satisfy \(P_{l}(1)=1\) for all values of \(l\)

Show that in the limit of low energies, the scattering phase shift for P-wave scattering by a hard sphere is proportional to \((k a)^{3}\) and therefore is negligible compared to the S-wave scattering phase shift. Hint. Use the asymptotic forms given in eqn \(14.32 \mathrm{c}\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free