The photoelectric current from Na (work function, w _ 0 = 2 . 3eV )...

Sodium metal :

$E = {E}_{0} + {\left(\text{ KE }\right)}$_\max ; {E}_{call}^{0} = 0 . 22V$</p><p>Cell reaction</p><p>Cathode :$AgCl\left(s\right) + {e}^{-} \to Ag\left(s\right) + {Cl}^{-}\left(aq\right)$</p><p>Anode :$\dfrac{1}{2}{H}_{2}\left(g\right) \to {H}^{+}\left(aq\right) + {e}^{-}$</p><p>Overall ;$AgCl\left(s\right) + \dfrac{1}{2}{H}_{2}\left(g\right) \to Ag\left(s\right) + {H}^{+}\left(aq\right) + {Cl}^{-}\left(aq\right)$</p><p>${E}_{cell} = {E}_{cell}^{0} - \dfrac{0 . 06}{1}\log \left[{H}^{+}\right]\left[{Cl}^{-}\right]$</p><p>${E}_{cell} = 0 . 22 - \dfrac{0 . 06}{1}\log \left[{10}^{- 1}\right]\left[{10}^{- 1}\right] = 0 . 22 + 0 . 12 = 0 . 34V$</p><p>${\left(KE\right)}_\max = {E}_{cell} = 0 . 34eV$</p><p>So$E = 2 . 3 + 0 . 34 = 2 . 64eV =$Energy of photon incident</p><p>For potassium metal :</p><p>$E = {E}_{0} + {\left(KE\right)}_\max$</p><p>$2 . 64 = 2 . 25 + {\left(KE\right)}_\max$</p><p>${\left(KE\right)}_\max = 0 . 39 = {E}_{cell}$</p><p>Cell reaction</p><p>Cathode :$AgCl ( s ) + {e}^{-} \to Ag\left(s\right) + {Cl}^{-}\left(aq\right)$</p><p>Anode :$\dfrac{1}{2}{H}_{2}\left(g\right) \to {H}^{+}\left(aq\right) + {e}^{-}$</p><p>Overall :$AgCl\left(s\right) + \dfrac{1}{2}{H}_{2}\left(g\right) \to Ag\left(s\right) + {H}^{+}\left(aq\right) + {Cl}^{-}\left(aq\right)$</p><p>${E}_{cell} = {E}_{cell}^{0} - \dfrac{0 . 06}{1}\log \left[{H}^{+}\right]\left[{Cl}^{-}\right]$</p><p>$0 . 39 = 0 . 22 - 0 . 12\log \left[{H}^{+}\right]$</p><p>$0 . 17 = 0 . 12 \times pH$</p><p>$pH = 17 / 12 = 1 . 4166 = 1 . 42