Correct Option: (C) Lyman series
Step 1 (Formula / Governing Law & Given Data): [1/3 Mark]
For photoelectric emission to occur, the energy of the incident photon ($E_{photon}$) must be greater than or equal to the work function ($\Phi_0$) of the metal surface:
$$E_{photon} \ge \Phi_0$$The work function of Zinc ($\Phi_0$) is approximately $4.3 \text{ eV}$. The energy of a photon emitted during a transition in a hydrogen atom is given by the Rydberg formula:
$$E = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \text{ eV}$$Step 2 (Step-by-Step Mathematical Working): [1/3 Mark]
- Lyman Series ($n_f = 1$): The minimum energy corresponds to the transition from $n_i = 2$ to $n_f = 1$: $$E_{min} = 13.6 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 13.6 \left( 1 - 0.25 \right) = 13.6 \times 0.75 = 10.2 \text{ eV}$$ Since $10.2 \text{ eV} > 4.3 \text{ eV}$, photons in the Lyman series have sufficient energy to eject photoelectrons.
- Balmer Series ($n_f = 2$): The maximum energy corresponds to the transition from $n_i = \infty$ to $n_f = 2$: $$E_{max} = 13.6 \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = \frac{13.6}{4} = 3.4 \text{ eV}$$ Since $3.4 \text{ eV} < 4.3 \text{ eV}$, Balmer series photons cannot eject photoelectrons.
- Paschen Series ($n_f = 3$): The energies are even lower than those of the Balmer series, hence they also fail to eject photoelectrons.
Step 3 (Final Concluded Answer): [1/3 Mark]
Since only the photons of the Lyman series possess energy greater than the work function of Zinc ($4.3 \text{ eV}$), they are the only ones capable of causing photoelectric emission.
💡 EL10 Concept Insight: Think of the work function as a "toll gate" fee; the Lyman series photons carry enough energy to pay this fee and still have energy left over to move, while the other series photons are simply too "weak" to get through the gate.