Discovering the Nucleus — Question 10

Section 42-1 · Physics pages 7010–7011

Question 10 A diffuse positive charge cannot produce the same short-range barrier Compare a point charge Ze with the same charge uniformly spread over a sphere of radius R. A projectile has charge q and K = qZe/(4πε0) > 0. Inside the sphere, take U(r) = K(3 −r2/R2)/(2R), with zero potential at infinity. r U R Schematic; labels, not drawing lengths, specify the data. (a) Find the largest potential barrier in the diffuse model and the condition for a head-on projectile to pass through its center. (b) For K = 144 MeV fm and atomic-scale R = 105 fm, compare the diffuse barrier with a 6.00 MeV projectile.
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Question 10 - Solution A diffuse positive charge cannot produce the same short-range barrier r U R (a) Method and calculation dU dr = −Kr/R3 ⇒Umax = U(0) = 3K/(2R) . E > 3K/(2R) permits passage through the center in this ideal model. (b) Interpretation and further calculation Umax = 216/105 = .00216 MeV = 2.16 keV . E/Umax = 6.00/.00216 = 2778 . Physical check Concentrating positive charge into a tiny nucleus allows a much larger Coulomb barrier than distributing it throughout the atom. This comparison isolates the central idea behind large-angle scattering evidence.

Question 10

A diffuse positive charge cannot produce the same short-range barrier

Compare a point charge Ze with the same charge uniformly spread over a sphere of radius R. A projectile has charge q and K=qZe/(4πε0)>0K=qZe/(4\pi\varepsilon_0)>0. Inside the sphere, take U(r)=K(3−r2/R2)/(2R)U(r)=K(3-r^2/R^2)/(2R), with zero potential at infinity.

Diagram shown in the original SVG worksheet.

  1. Find the largest potential barrier in the diffuse model and the condition for a head-on projectile to pass through its center.

  2. For K=144MeVfmK=144\,\mathrm{MeV\,fm} and atomic-scale R=105fmR=10^5\,\mathrm{fm}, compare the diffuse barrier with a 6.00 MeV projectile.

Question 10 - Solution

A diffuse positive charge cannot produce the same short-range barrier

Diagram shown in the original SVG worksheet.

(a) Method and calculation

dUdr=−Kr/R3⇒Umax=U(0)=3K/(2R).\frac{dU}{dr}=-Kr/R^3\Rightarrow\boxed{U_{\max}=U(0)=3K/(2R)}.E>3K/(2R)permits passage through the center in this ideal model.\boxed{E>3K/(2R)}\quad\text{permits passage through the center in this ideal model}.

(b) Interpretation and further calculation

Umax=216/105=.00216MeV=2.16keV.\boxed{U_{\max}=216/10^5=.00216\,\mathrm{MeV}=2.16\,\mathrm{keV}}.E/Umax=6.00/.00216=2778.\boxed{E/U_{\max}=6.00/.00216=2778}.

Physical check

Concentrating positive charge into a tiny nucleus allows a much larger Coulomb barrier than distributing it throughout the atom. This comparison isolates the central idea behind large-angle scattering evidence.