Talk:Neutron scattering lengths

Origin of the scattering lengths

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

Neutron energy

Consider first the energies of neutrons used in scattering experiments (recall the neutron mass is 1.67×10⁻²⁷ kg). A thermal neutron (~100°C) would have energy of:

$\mathrm {KE} ={\frac {1}{2}}mv^{2}={\frac {3}{2}}kT=7\times 10^{-21}\,\mathrm {J} =48\,\mathrm {meV}$

The velocity of such neutrons is ~3000 m/s, and the momentum is $p=mv=5\times 10^{-24}\,\mathrm {kgm/s}$. Finally, the deBroglie wavelength would be:

$\lambda ={\frac {h}{p}}=1.3\,\mathrm {\AA}$

A cold neutron (~18 K) would have energy of 4×10⁻²² J = 2 meV, velocity of ~660 m/s, and wavelength of 6 Å.

Potential well

Consider a neutron of energy $E_{i}$ 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 $kR={\sqrt {2mE_{i}}}R2\pi /h\ll 1$; because of the small neutron mass and energy (see above), as well as the small size of a nucleus (femtometers). Therefore:

${\begin{alignedat}{2}\psi _{s,\mathrm {out} }&={\frac {\sin(kR)}{kR}}-b{\frac {e^{ikR}}{r}}\\&\approx 1-b/r\end{alignedat}}$

and so:

${\begin{alignedat}{2}\psi _{s,\mathrm {out} }(r=R)&=\psi _{s,\mathrm {in} }(r=R)\\1-{\frac {b}{R}}&=A{\frac {\sin(qR)}{qR}}\\R-b&=A{\frac {\sin(qR)}{q}}\end{alignedat}}$

And from equating the derivatives:

${\begin{alignedat}{2}1&=A\cos(qR)\\A&={\frac {1}{\cos(qR)}}\end{alignedat}}$