Atoms and Nuclei

$\because$ For a hydrogen-like atom/ion, $E_n \propto \dfrac{Z^2}{n^2}$, so the energy of radiation emitted in electronic transitions is $E \propto Z^2\left(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right)$.$\Rightarrow (P) \longrightarrow (5)$$\because$ Moseley's law for characteristic $x$-rays gives $\sqrt{\nu}=a(Z-b)$, and for the $K_\alpha$ line the screening constant is $b \approx 1$, hence $E=h\nu \propto (Z-1)^2$.$\Rightarrow (Q) \longrightarrow (1)$$\because$ The electrostatic (Coulomb) part of the nuclear binding energy is due to mutual repulsion of the $Z$ protons, so it is proportional to the number of proton pairs, $E_C \propto Z(Z-1)$ (with $R \propto A^{1/3}$).$\Rightarrow (R) \longrightarrow (2)$$\because$ For stable nuclei with mass numbers in the range 30 to 170, the average binding energy per nucleon is nearly constant ($\approx 8\ \text{MeV}$), i.e. practically independent of $Z$.$\Rightarrow (S) \longrightarrow (4)$$\therefore$ Correct match: $P \to 5$, $Q \to 1$, $R \to 2$, $S \to 4$ (entry (3), the energy of continuous $x$-rays, remains unmatched).