Some Nuclear Properties — Question 10

Section 42-2 · Physics pages 7030–7031

Question 10 A nuclear quadrupole moment distinguishes shape from volume For a uniform charge ellipsoid with semiaxes a,a,c and total charge Ze, use ⟨x2⟩= ⟨y2⟩= a2/5 and ⟨z2⟩= c2/5. Define intrinsic quadrupole moment Q0 = ∫ρq(3z2 −r2)dV . a c Schematic; labels, not drawing lengths, specify the data. (a) Derive Q0 and determine its sign for prolate and oblate shapes. (b) For Z = 40, a = 4.00 fm and c = 5.00 fm, calculate Q0/e and the radius of an equal-volume sphere.
Show solutionHide solution
Question 10 - Solution A nuclear quadrupole moment distinguishes shape from volume a c (a) Method and calculation Q0 = Ze[2⟨z2⟩−⟨x2⟩−⟨y2⟩] = 2Ze 5 (c2 −a2) . c > a : Q0 > 0; c < a : Q0 < 0; c = a : Q0 = 0 . (b) Interpretation and further calculation Q0/e = (2 × 40/5)(25 −16) = 144 fm2 . 4πR3 3 = 4πa2c 3 ⇒R = (a2c)1/3 = 4.309 fm . Physical check This is an intrinsic, body-fixed shape moment. The observed laboratory quadrupole moment also depends on the nuclear angular-momentum state.

Question 10

A nuclear quadrupole moment distinguishes shape from volume

For a uniform charge ellipsoid with semiaxes a,a,c and total charge Ze, use ⟨x2⟩=⟨y2⟩=a2/5\langle x^2\rangle=\langle y^2\rangle=a^2/5 and ⟨z2⟩=c2/5\langle z^2\rangle=c^2/5. Define intrinsic quadrupole moment Q0=∫ρq(3z2−r2)dVQ_0=\int\rho_q(3z^2-r^2)dV.

Diagram shown in the original SVG worksheet.

  1. Derive Q0Q_0 and determine its sign for prolate and oblate shapes.

  2. For Z=40Z=40, a=4.00fma=4.00\,\mathrm{fm} and c=5.00fmc=5.00\,\mathrm{fm}, calculate Q0/eQ_0/e and the radius of an equal-volume sphere.

Question 10 - Solution

A nuclear quadrupole moment distinguishes shape from volume

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Q0=Ze[2⟨z2⟩−⟨x2⟩−⟨y2⟩]=2Ze5(c2−a2).Q_0=Ze[2\langle z^2\rangle-\langle x^2\rangle-\langle y^2\rangle]=\boxed{\frac{2Ze}{5}(c^2-a^2)}.c>a:Q0>0;c<a:Q0<0;c=a:Q0=0.\boxed{c>a:\ Q_0>0;\qquad c<a:\ Q_0<0;\qquad c=a:\ Q_0=0}.

(b) Interpretation and further calculation

Q0/e=(2×40/5)(25−16)=144fm2.\boxed{Q_0/e=(2\times40/5)(25-16)=144\,\mathrm{fm^2}}.4πR33=4πa2c3⇒R=(a2c)1/3=4.309fm.\frac{4\pi R^3}{3}=\frac{4\pi a^2c}{3}\Rightarrow\boxed{R=(a^2c)^{1/3}=4.309\,\mathrm{fm}}.

Physical check

This is an intrinsic, body-fixed shape moment. The observed laboratory quadrupole moment also depends on the nuclear angular-momentum state.