= Solution
\b[Closest approach.] In the heavy-target approximation, a head-on <alpha particle> of charge $2e$ converts its initial <kinetic energy> to repulsive <Coulomb potential> energy at its turning point. The <nuclear radius bound from a Coulomb turning point> is therefore
$$
E_\alpha=\frac{2Ze^2}{4\pi\varepsilon_0r_{\min}},\qquad
r_{\min}=\frac{2(82)(1.440\ \mathrm{MeV\,fm})}{5.51\ \mathrm{MeV}}.
$$
Thus the point-projectile interpretation of the radius bound is
$$
\boxed{R_{\rm Pb}\lesssim4.3\times10^{-14}\ \mathrm m.}
$$
It is an upper limit from Coulomb scattering, not a measurement of the actual nuclear radius. If “size” denotes diameter, the associated bound is twice this value. Including <nuclear recoil>, $E_{\rm cm}\simeq[207/(207+4)]E_\alpha$, increases the closest separation by about two per cent to $4.4\times10^{-14}\ \mathrm m$.
\b[Count-rate estimate and normalization.] The graph has approximate slope $\Delta\ln S/\Delta\ln A\simeq15/10=3/2$. It therefore supplies the empirical <log-log plot> relation $S\propto A^{3/2}$. Calibrating its normalization with the lead observation gives
$$
\boxed{S_{\rm Ag}\simeq60\left(\frac{108}{207}\right)^{3/2}
\simeq23\ \text{particles per minute}.}
$$
The <Rutherford differential cross-section> at fixed beam energy and scattering angle instead has a $Z^2$ dependence per target nucleus. More generally, the <target normalization in Rutherford scattering rates> is
$$
\frac{S_{\rm Ag}}{S_{\rm Pb}}
=\frac{\mathcal N_{\rm Ag}}{\mathcal N_{\rm Pb}}
\left(\frac{47}{82}\right)^2,
$$
where $\mathcal N$ is the target nuclei per unit area. If the foils have equal $\mathcal N$, the theoretical estimate is $19.7$, or about $20$ particles per minute, close to the graphical estimate at its rough level. Equal foil mass per area would instead give $37.8$ per minute because $\mathcal N\propto1/A$.
The source does not specify that target normalization. Thus $23$ per minute is the estimate from its supplied graph; an exact rate prediction from Rutherford theory alone is underdetermined without comparable foil areal densities. The empirical $A^{3/2}$ slope and an exact equal-$\mathcal N$ $Z^2$ law should not be presented as identical assumptions.
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