Elastic scattering of electrons from a light nucleus is measured at two small angles

PHYSICS · Multi-part · Fermi University Physics · Nuclear Physics · Nuclear constituents, size and binding · Nuclear constituents, size and binding

The Nucleus Resolved, Ch How big a nucleus is, and what the mass surface knows

Problem

Elastic scattering of

300\,\mathrm{MeV}

electrons from a light nucleus is measured at two small angles, and the form factor is extracted at each. The results are

F(q_1^2) = 0.9494

\theta_1 = 6^\circ

and

F(q_2^2) = 0.7980

\theta_2 = 12^\circ

. Someone sets out to find the rms charge radius from the small-

q

expansion

F(q^2) = 1 - q^2\langle r^2\rangle/6

, and writes: "The momentum transfer at angle

\theta

q = 2k\sin\theta

, with

k = E/\hbar c

. For

E = 300\,\mathrm{MeV}

k = 300/197.327 = 1.520\,\mathrm{fm^{-1}}

. So

q_1 = 2(1.520)\sin 6^\circ = 0.318\,\mathrm{fm^{-1}}

and

q_2 = 2(1.520)\sin 12^\circ = 0.632\,\mathrm{fm^{-1}}

. The slope is

\frac{F(q_2^2) - F(q_1^2)}{q_2^2 - q_1^2} = \frac{0.7980 - 0.9494}{0.3996 - 0.1010} = -0.507\,\mathrm{fm^2}.

Since

F = 1 - q^2\langle r^2\rangle /6

\langle r^2\rangle = -6\times(-0.507) = 3.04\,\mathrm{fm^2}

, so

\langle r^2\rangle^{1/2} = 1.74\,\mathrm{fm}

." ((a)) Find the error in this working. ((b)) Keeping the same leading-order small-

q

model, with higher powers of

q

neglected, give the corrected estimate of

\langle r^2\rangle^{1/2}

for this nucleus.