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In an experiment, an ammeter is connected in series with an evacuated tube containing two electrodes (A and B). This combination is connected in parallel with a potentiometer wire that has a cell of emf 10 V in series. Electrode B is illuminated by light comprising four wavelengths: 155 nm, 310 nm, 620 nm, and 1220 nm. The work function of electrode B is 1.4 eV. The jockey is kept at point Q, with PQ = l cm (total wire length = 100 cm). Which of the following statements are correct?

1. For l = 60 cm, the maximum kinetic energy of a photoelectron striking electrode A is 0.6 eV
2. For l = 80 cm, the ammeter shows no deflection
3. For l = 42 cm, the ammeter deflection is greater than for l = 60 cm, but the maximum KE of photoelectrons striking A is four times that at l = 60 cm
4. If the terminals of the cell are interchanged, the maximum KE of the electrons striking the electrode becomes 18 eV

Correct answer: For l = 60 cm, the maximum kinetic energy of a photoelectron striking electrode A is 0.6 eV

Solution

Photon energies: 155 nm -> 1240/155 = 8 eV; 310 nm -> 1240/310 = 4 eV; 620 nm -> 2 eV (< 1.4 eV... actually 2 > 1.4, so emission occurs); 1220 nm -> ~1.02 eV (no emission). Wait: work function = 1.4 eV. So 620 nm (2 eV) also gives emission with KE_max = 0.6 eV. The highest KE comes from 155 nm: KE_max = 8 - 1.4 = 6.6 eV. At l = 60 cm: retarding potential = 6 V. Electrons from 155 nm that reach A have KE = 6.6 - 6.0 = 0.6 eV. Statement A is correct. At l = 80 cm: retarding V = 8 V > 6.6 eV, so no electrons reach A — ammeter shows no deflection (B is correct). At l = 42 cm: accelerating V =... wait, if jockey at Q and current flows, V_PQ = 4.2 V. If this is a retarding potential, electrons from 155 nm reach A with KE = 6.6 - 4.2 = 2.4 eV = 4 * 0.6 eV. More electrons reach A (higher current), consistent with C. So A, B, C are correct. For D: if terminals reversed, V = 10 V added to KE: 6.6 + 10 = 16.6 eV, not 18 eV. D is incorrect.

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