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 pH of aq. HCl required to stop the photoelectric current from K($$w_0 = 2.25$$ eV), all other conditions remaining the same, is ......... $$\times 10^{-2}$$ (to the nearest integer).
Given $$2.303\frac{RT}{F} = 0.06$$ V; $$E^0_{AgCl/Ag/Cl} = 0.22$$ V

We begin by recalling the Einstein photo-electric equation. When monochromatic light of energy $$h\nu$$ falls on a metal of work function $$w_0$$, the maximum kinetic energy of the photo-electrons is

$$ \dfrac{1}{2}mv^{2}_{\max}=h\nu-w_0 $$

If the electrons are just stopped by a retarding (stopping) potential $$V_s$$, then each electron of charge $$e$$ loses an energy $$eV_s$$, so

$$ eV_s=h\nu-w_0\quad\Longrightarrow\quad V_s=\dfrac{h\nu}{e}-w_0/e. $$

Thus, for the same light, the change in the stopping potential is exactly the change in the work function expressed in electron-volts, because $$1\ \text{eV}=1\ \text{V}\cdot e$$ for a single electron:

$$\Delta V_s=\Delta w_0\ (\text{in eV}).$$

For Na and K we have

$$ w_0(\text{Na})=2.30\ \text{eV},\qquad w_0(\text{K})=2.25\ \text{eV}. $$

Therefore

$$ \Delta w_0=w_0(\text{Na})-w_0(\text{K})=2.30-2.25=0.05\ \text{eV}, $$

so the stopping potential for K must be larger by

$$ \Delta V_s = 0.05\ \text{V}. $$

Next, we analyze the electrochemical cell that supplies the stopping potential:

$$ \text{Pt(s)}\vert\text{H}_2(g,1\ \text{bar})\vert\text{HCl(aq, pH}=1)\vert\text{AgCl(s)}\vert\text{Ag(s)}. $$

The left-hand electrode is the hydrogen electrode, the right-hand electrode is the AgCl/Ag electrode. All electrode potentials will be written as reduction potentials.

(i) Reduction potential of the $$\text{H}^+/\text{H}_2$$ couple

For the reduction

$$ 2\text{H}^+ + 2e^- \rightarrow \text{H}_2(g) $$

the Nernst equation is

$$ E_{\text{H}^+/\text{H}_2} = E^{0} + 0.06\log[\text{H}^+],\qquad E^{0}=0.00\ \text{V}. $$

At $$\text{pH}=1$$ we have $$[\text{H}^+]=10^{-1}\ \text{M}$$, so

$$ \log[\text{H}^+] = \log(10^{-1}) = -1, $$

and therefore

$$ E_{\text{H}^+/\text{H}_2} = 0 + 0.06(-1) = -0.06\ \text{V}. $$

(ii) Reduction potential of the $$\text{AgCl/Ag/Cl}^-$$ couple

The reduction is

$$ \text{AgCl(s)} + e^- \rightarrow \text{Ag(s)} + \text{Cl}^- (\text{aq}). $$

The Nernst equation is

$$ E_{\text{AgCl/Ag}} = E^{0}_{\text{AgCl/Ag}} - 0.06\log[\text{Cl}^-],\qquad E^{0}_{\text{AgCl/Ag}} = 0.22\ \text{V}. $$

Because the electrolyte is HCl and the acid is strong and completely dissociated, $$[\text{Cl}^-]=[\text{H}^+]=10^{-1}\ \text{M}$$ at pH = 1. Hence

$$ \log[\text{Cl}^-] = -1,\qquad E_{\text{AgCl/Ag}} = 0.22 - 0.06(-1) = 0.22 + 0.06 = 0.28\ \text{V}. $$

(iii) Emf of the cell at pH = 1

The cell emf is

$$ E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} = E_{\text{AgCl/Ag}} - E_{\text{H}^+/\text{H}_2} = 0.28\ \text{V} - (-0.06\ \text{V}) = 0.34\ \text{V}. $$

This 0.34 V exactly balances the stopping potential for Na, so

$$ V_s(\text{Na}) = 0.34\ \text{V}. $$

(iv) Stopping potential required for K

Since the work function of K is 0.05 eV lower,

$$ V_s(\text{K}) = V_s(\text{Na}) + 0.05\ \text{V} = 0.34 + 0.05 = 0.39\ \text{V}. $$

(v) Required emf of the cell at unknown pH = $$x$$

Let the new pH be $$x$$. Then

$$ [\text{H}^+] = 10^{-x},\qquad [\text{Cl}^-] = 10^{-x}. $$

The reduction potentials now become

$$ E_{\text{H}^+/\text{H}_2} = 0.06\log[\text{H}^+] = 0.06\log(10^{-x}) = 0.06(-x) = -0.06x\ \text{V}, $$

$$ E_{\text{AgCl/Ag}} = 0.22 - 0.06\log[\text{Cl}^-] = 0.22 - 0.06\log(10^{-x}) = 0.22 - 0.06(-x) = 0.22 + 0.06x\ \text{V}. $$

Therefore the emf is

$$ \begin{aligned} E_{\text{cell}} &= E_{\text{AgCl/Ag}} - E_{\text{H}^+/\text{H}_2} \\ &= (0.22 + 0.06x) - (-0.06x) \\ &= 0.22 + 0.12x\ \text{V}. \end{aligned} $$

This emf must equal the required stopping potential of 0.39 V:

$$ 0.22 + 0.12x = 0.39. $$

Solving for $$x$$,

$$ 0.12x = 0.39 - 0.22 = 0.17, $$

$$ x = \dfrac{0.17}{0.12} = 1.416\ (\text{approximately}). $$

So the required pH is

$$ \text{pH} \;=\; 1.42\quad(\text{to two decimal places}). $$

The question asks us to express the answer as “……. $$\times 10^{-2}$$ (to the nearest integer)”. Writing

$$ 1.42 = 142 \times 10^{-2}, $$

the integer sought is 142.

Hence, the correct answer is Option 142.