Question 16

(P) According to Bohr's model, the energy of electronic transitions in hydrogen-like atoms is given by $E = 13.6 Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$, so $E \propto Z^2$. Thus, P $\rightarrow$ 5.

(Q) According to Moseley's Law for characteristic X-rays, the frequency $\nu$ is related to $Z$ as $\sqrt{\nu} \propto (Z - b)$. For $K_\alpha$ lines, $b \approx 1$. Since $E = h\nu$, $E \propto (Z - 1)^2$. Thus, Q $\rightarrow$ 1.

(R) The electrostatic (Coulomb) potential energy of a nucleus with $Z$ protons is proportional to the number of proton pairs, which is $\frac{Z(Z-1)}{2}$. Thus, the electrostatic part of binding energy $E \propto Z(Z-1)$. Thus, R $\rightarrow$ 2.

(S) For stable nuclei with mass numbers $30 < A < 170$, the binding energy per nucleon ($BE/A$) is nearly constant (approximately $8$ MeV/nucleon). Since $A$ is roughly proportional to $Z$ for stable nuclei, $E_{avg}$ is practically independent of $Z$. Thus, S $\rightarrow$ 4.

Matching these results: P-5, Q-1, R-2, S-4. Correct Option is C.