Talk:Neutron scattering lengths

Origin of the scattering lengths

The following description is adapted from Boualem Hammouda's (NCNR) SANS tutorial.

Consider first the energies of neutrons used in scattering experiments. A thermal neutron , the energy for even a thermal neutron (1.8 Å wavelength) is

Consider a neutron of energy Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_i} interacting with a nucleus, which exhibits an attractive square well of depth $-V_{0}$ and width $2R$; where $V_{0}\gg E_{i}$. The Schrödinger equation is:

$\left[-{\frac {h^{2}}{8\pi ^{2}m}}\nabla ^{2}+V(r)\right]\psi (r)=E\psi (r)$

Outside of the square-well ($|r|>R$), $V(r)=0$, and so the equation is solved as simply:

$\psi _{s,\mathrm {out} }={\frac {\sin(kr)}{kr}}-b{\frac {e^{ikr}}{r}}$

where $k={\sqrt {2mE_{i}}}2\pi /h$. Inside the square-well ($|r|<R$), the potential is $V(r)=-V_{0}$, and the solution becomes:

$\psi _{s,\mathrm {in} }=A{\frac {\sin(qr)}{qr}}$

where $q={\sqrt {2m(E_{i}+V_{0})}}2\pi /h$. The two solutions are subject to a continuity boundary condition at $|r|=R$:

${\begin{alignedat}{2}\psi _{s,\mathrm {out} }(r=R)&=\psi _{s,\mathrm {in} }(r=R)\\{\frac {\mathrm {d} }{\mathrm {d} r}}\psi _{s,\mathrm {out} }(r=R)&={\frac {\mathrm {d} }{\mathrm {d} r}}\psi _{s,\mathrm {in} }(r=R)\end{alignedat}}$

Note that the mass of a neutron is ~10⁻²⁷ kg

Note that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle kR = \sqrt{2 m E_i} R 2\pi/h \ll 1} for , and