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:

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 \mathrm{KE} = \frac{1}{2} m v^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 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 p=mv=5\times10^{-24} \, \mathrm{kgm/s}} . Finally, the deBroglie wavelength would be:

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 \lambda = \frac{h}{p} = 1.3 \, \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 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 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 -V_0} and width 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 2R} ; where 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 V_0 \gg E_i} . The Schrödinger equation is:

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

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

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

where 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 k=\sqrt{2mE_i} 2 \pi/h} . Inside the square-well (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 |r|<R} ), the potential is 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 V(r)=-V_0} , and the solution becomes:

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 \psi_{s,\mathrm{in}} = A \frac{\sin(qr)}{qr} }

where 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 q=\sqrt{2m(E_i+V_0)} 2 \pi/h} . The two solutions are subject to a continuity boundary condition at 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 |r|=R} :

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 \begin{alignat}{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{alignat} }

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} ; because of the small neutron mass and energy (see above), as well as the small size of a nucleus (femtometers). Therefore:

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

and so:

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 \begin{alignat}{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{alignat} }

And from equating the derivatives:

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