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JEE Main 2020 Chemistry question

ChemistryElectrochemistryJEE Main 2020

The photoelectric current from Na (work function, = 2.3 eV) is stopped by the output voltage of the cell Pt(s) | (g, 1 bar) | HCl (aq., pH = 1) | AgCl(s) | Ag(s). The pH of aq. HCl required to stop the photoelectric current from K( = 2.25 eV), all other conditions remaining the same, is × (to the nearest integer). Given = 0.06 V ; = 0.22 V

Answer: 142

Solution

The cell voltage at pH 1 is 0.22 + 0.12 = 0.34 V, so the photon energy is 2.3 + 0.34 = 2.64 eV; for K the stopping potential is 2.64 − 2.25 = 0.39 V =

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Question ID VK-012271