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| − | ==Origin of the scattering lengths==
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| − | The following description is adapted from [http://www.ncnr.nist.gov/programs/sans/pdf/polymer_tut.pdf Boualem Hammouda's (NCNR) SANS tutorial].
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| − | ===Neutron energy===
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| − | Consider first the energies of neutrons used in scattering experiments (recall the neutron mass is 1.67×10<sup>−27</sup> kg). A thermal neutron (~100°C) would have energy of:
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| − | :<math>
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| − | \mathrm{KE} = \frac{1}{2} m v^2 = \frac{3}{2} kT = 7 \times 10^{-21} \, \mathrm{J} = 48 \,\mathrm{meV}
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| − | </math>
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| − | The velocity of such neutrons is ~3000 m/s, and the momentum is <math>p=mv=5\times10^{-24} \, \mathrm{kgm/s}</math>. Finally, the deBroglie wavelength would be:
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| − | :<math>
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| − | \lambda = \frac{h}{p} = 1.3 \, \AA
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| − | </math>
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| − | A cold neutron (~18 K) would have energy of 4×10<sup>−22</sup> J = 2 meV, velocity of ~660 m/s, and wavelength of 6 Å.
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| − | ===Potential well===
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| − | Consider a neutron of energy <math>E_i</math> interacting with a nucleus, which exhibits an attractive square well of depth <math>-V_0</math> and width <math>2R</math>; where <math>V_0 \gg E_i</math>. The [http://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation Schrödinger equation] is:
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| − | :<math>
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| − | \left[ - \frac{h^2}{8 \pi^2 m}\nabla^2 + V(r) \right] \psi(r) = E \psi(r)
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| − | </math>
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| − | Outside of the square-well (<math>|r|>R</math>), <math>V(r)=0</math>, and so the equation is solved as simply:
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| − | :<math>
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| − | \psi_{s,\mathrm{out}} = \frac{\sin(kr)}{kr} - b \frac{e^{ikr}}{r}
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| − | </math>
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| − | where <math>k=\sqrt{2mE_i} 2 \pi/h</math>. Inside the square-well (<math>|r|<R</math>), the potential is <math>V(r)=-V_0</math>, and the solution becomes:
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| − | :<math>
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| − | \psi_{s,\mathrm{in}} = A \frac{\sin(qr)}{qr}
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| − | </math>
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| − | where <math>q=\sqrt{2m(E_i+V_0)} 2 \pi/h</math>. The two solutions are subject to a continuity boundary condition at <math>|r|=R</math>:
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| − | :<math>\begin{alignat}{2}
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| − | \psi_{s,\mathrm{out}} (r=R) & = \psi_{s,\mathrm{in}} (r=R) \\
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| − | \frac{\mathrm{d}}{\mathrm{d}r} \psi_{s,\mathrm{out}} (r=R) & = \frac{\mathrm{d}}{\mathrm{d}r} \psi_{s,\mathrm{in}} (r=R)
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| − | \end{alignat}
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| − | </math>
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| − | Note that <math>kR = \sqrt{2 m E_i} R 2\pi/h \ll 1</math>; because of the small neutron mass and energy (see above), as well as the small size of a nucleus (femtometers). Therefore:
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| − | :<math>
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| − | \begin{alignat}{2}
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| − | \psi_{s,\mathrm{out}} & = \frac{\sin(kR)}{kR} - b \frac{e^{ikR}}{r} \\
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| − | & \approx 1 - b/r
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| − | \end{alignat}
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| − | </math>
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| − | and so:
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| − | :<math>
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| − | \begin{alignat}{2}
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| − | \psi_{s,\mathrm{out}} (r=R) & = \psi_{s,\mathrm{in}} (r=R) \\
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| − | 1 - \frac{b}{R} & = A \frac{\sin(qR)}{qR} \\
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| − | R - b & = A \frac{\sin(qR)}{q}
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| − | \end{alignat}
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| − | </math>
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| − | And from equating the derivatives:
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| − | :<math>
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| − | \begin{alignat}{2}
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| − | 1 & = A \cos(qR) \\
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| − | A & = \frac{1}{ \cos(qR) }
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| − | \end{alignat}
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| − | </math>
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