Q4. Find the shortest wavelength of the Paschen series (R = 1.097 × 107 m−1).
For ni → ∞ and nf = 3, 1/λ = R/9, so λ = 9/R ≈ 8.20 × 10−7 m = 820 nm, in the infrared. NCERT Figure 12.5 (page 297) marks this limit; the formula is beyond the NCERT text.
Q5. Can hydrogen in its ground state absorb a photon of 12.4 eV? Of 13.06 eV?
12.4 eV: no. It lies between the gaps to n = 3 (12.09 eV) and n = 4 (12.75 eV), so it matches no gap, and it is too small to free the electron, which needs 13.6 eV. 13.06 eV: at the rounded precision of the question, yes; the ideal gap is 13.056 eV, since 13.6 × (1 − 1/25) ≈ 13.06 eV, it lifts the atom from n = 1 to n = 5 (reasoned from NCERT pages 300–301).
Q6. A 15 eV photon falls on a hydrogen atom in the ground state. If it is absorbed in photoionisation, what is the emitted electron’s kinetic energy, neglecting recoil?
The absorbed photon frees the electron, which needs 13.6 eV, and the remaining 15 − 13.6 = 1.4 eV becomes the electron’s kinetic energy (reasoned from NCERT page 300).
Q7. Find the ratio of the longest wavelength of the Balmer series to the longest wavelength of the Lyman series.
Balmer, 3 → 2: 1/λ = R(1/4 − 1/9) = 5R/36. Lyman, 2 → 1: 1/λ = R(1 − 1/4) = 3R/4. The ratio is (36/5)/(4/3) = 27/5 = 5.4, that is 656 nm against 122 nm (beyond the NCERT text).
Q8. Which photon has the higher frequency: the one from n = 3 to n = 2, or the one from n = 2 to n = 1?
The 2 → 1 photon: it carries 10.2 eV, against 1.89 eV for 3 → 2, and ν = E/h (NCERT Eq. (12.11), page 301; energies from NCERT page 300).