The photoelectric current from Na (work function, w 0 = 2 . 3 eV ) is stopped by the output voltage of the cell Pt ( s ) ‖ H 2 ( g , 1 bar ) | HCl ( aq · , pH = 1 ) | AgCl ( s ) ∣ Ag ( s ) the p H of aq. HCl required to stop the photoelectric current from K w 0 = 2 . 25 eV , all other conditions remaining the same, is … … … . × 10 - 2 (to the nearest integer). Given 2 . 303 RT

Step-by-step solution

Sodium metal : E = E 0 + KE max ;   E call 0 = 0 . 22   V Cell reaction Cathode : AgCl s + e -   →   Ag s + Cl - aq Anode : 1 2   H 2 g   →   H + aq + e - Overall ; AgCl s + 1 2 H 2 g   →   Ag s + H + aq + Cl - aq E cell = E cell 0 - 0 . 06 1 log H +   Cl - E cell = 0 . 22 - 0 . 06 1 log 10 - 1 10 - 1 = 0 . 22 + 0 . 12 = 0 . 34   V KE max = E cell = 0 . 34   eV So E = 2 . 3 + 0 . 34 = 2 . 64   eV = Energy of photon incident For potas