Structure of Atom, on the exam screen.

The orbital is named by n and the subshell letter for l (l = 0,1,2,3 -> s,p,d,f). So n=2,l=1 is 2p (II); n=4,l=0 is 4s (III); n=5,l=3 is 5f (IV); n=3,l=2 is 3d (I). This gives P-II, Q-III, R-IV, S-I.

In one second the bulb consumes 150 J; only 8% becomes light, so light energy per second = $150 \times 0.08 = 12$ J. The number of photons n = (light energy)/(energy per photon) = $12 / (4.42 \times 10^{-19}) \approx 2.71 \times 10^{19}$. The other options come from omitting the 8% factor or arithmetic errors.

A radial node is a distance at which the wavefunction actually CHANGES SIGN, so on a $\psi_r$ versus r plot you simply count how many times the curve crosses the horizontal axis. Merely approaching zero asymptotically at large r does not count. The plot labelled C crosses the axis twice - once into negative values and once back up - so it has two radial nodes, matching an orbital with $n - l - 1 = 2$ such as 3s. NCERT Class 11, Structure of Atom (quantum mechanical model, nodes).

Why the other options are wrong:

- (1) shows a curve that decays without ever changing sign - zero radial nodes, as in 1s
- (2) crosses the axis only once, so it has a single radial node
- (4) crosses more often than twice, so its node count is higher than asked for

The atomic number (71) gives the number of protons and electrons, and the mass number (175) minus the atomic number gives the number of neutrons: 175 - 71 = 104. Chapter: Structure of Atom.