This question requires matching the functional dependencies of energy (E) on the atomic number (Z) in List-I with the corresponding physical phenomena in List-II.

Step 1: Analyze (P) $E \propto Z^2$

This dependence is a hallmark of the energy levels in a hydrogen-like atom (an atom with a single electron and a nucleus with charge $+Ze$). According to the Bohr model, the energy of the electron in the n-th stationary orbit is given by: $E_n = -\frac{13.6 \text{ eV}}{n^2} Z^2$ Thus, the energy of any particular state is proportional to $Z^2$. The energy of radiation emitted during an electronic transition between two states, say $n_2$ and $n_1$, is also proportional to $Z^2$: $\Delta E = E_{n_2} - E_{n_1} = 13.6 Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$ This matches with List-II, item (5): "energy of radiation due to electronic transitions from hydrogen-like atoms". Therefore, P → 5.

Step 2: Analyze (Q) $E \propto (Z - 1)^2$

This dependence is described by Moseley's Law for characteristic X-rays. When an electron is knocked out from an inner shell (like the K-shell, n=1), an electron from a higher shell (like the L-shell, n=2) fills the vacancy, emitting an X-ray photon. The transitioning electron experiences a nuclear charge that is screened by the other electrons. For a K-alpha transition (L→K), the transitioning electron is shielded by the one remaining electron in the K-shell. Thus, the effective nuclear charge it experiences is approximately $(Z-1)e$. The energy of the emitted X-ray is then given by a formula similar to the hydrogen atom transition, but with $Z$ replaced by the effective charge $(Z-b)$, where $b$ is the screening constant. For K-series X-rays, $b \approx 1$. $E \approx 13.6 (Z-1)^2 \left( \frac{1}{1^2} - \frac{1}{n^2} \right), \quad n=2,3,...$ This matches with List-II, item (1): "energy of characteristic x-rays". Therefore, Q → 1.

Step 3: Analyze (R) $E \propto Z(Z - 1)$

This term relates to the electrostatic potential energy of the protons within a nucleus. A nucleus with atomic number $Z$ contains $Z$ protons. The electrostatic repulsion exists between every pair of protons. The number of such pairs is given by the combination formula $\binom{Z}{2} = \frac{Z(Z-1)}{2}$. The total Coulomb repulsion energy, which reduces the nuclear binding energy, is proportional to this number of pairs. $E_c \propto \frac{Z(Z-1)}{2} \frac{e^2}{r_{avg}}$ This electrostatic energy is a key component of the semi-empirical mass formula. This matches with List-II, item (2): "electrostatic part of the nuclear binding energy...". Therefore, R → 2.

Step 4: Analyze (S) $E$ is practically independent of $Z$

This statement describes the behavior of the average binding energy per nucleon (B.E./A) for stable nuclei over a wide range of mass numbers, specifically from A=30 to A=170. In this region, the B.E./A curve is relatively flat, at a nearly constant value of about 8.5 MeV per nucleon. This phenomenon, known as the saturation of nuclear forces, implies that the average binding energy per nucleon is largely independent of the total number of nucleons (A) and the number of protons (Z) for these nuclei. This matches with List-II, item (4): "average nuclear binding energy per nucleon for stable nuclei with mass number in the range 30 to 170". Therefore, S → 4.

Step 5: Consolidate the matches and choose the correct option

Based on the analysis:

• P → 5
• Q → 1
• R → 2
• S → 4

This set of matches corresponds to option C.

• (P) $E \propto Z^2$ → (5) energy of radiation due to electronic transitions from hydrogen-like atoms
• (Q) $E \propto (Z - 1)^2$ → (1) energy of characteristic x-rays
• (R) $E \propto Z(Z - 1)$ → (2) electrostatic part of the nuclear binding energy
• (S) $E$ is practically independent of $Z$ → (4) average nuclear binding energy per nucleon

The correct option is C: P→5, Q→1, R→2, S→4.