1. Photoelectric stopping condition

For a metal, the maximum kinetic energy of emitted electrons is $K_{\max}=h\nu-w_0$ If the stopping potential is $V_s$, then $eV_s=K_{\max}=h\nu-w_0$

For the same incident light, $h\nu$ is unchanged. Hence: $eV_{s,Na}=h\nu-2.30$ $eV_{s,K}=h\nu-2.25$ Subtracting, $eV_{s,K}-eV_{s,Na}=2.30-2.25=0.05\ \text{eV}$ Since $1\ \text{eV}=e\times 1\ \text{V}$, $V_{s,K}-V_{s,Na}=0.05\ \text{V}$ So the stopping potential needed for $K$ is larger by $0.05\,$V.

2. EMF of the given cell

Cell: $\text{Pt}|H_2(1\,\text{bar})|HCl(\text{aq},\,\text{pH}=1)|AgCl(s)|Ag(s)$

Left electrode is hydrogen electrode: $2H^+ +2e^- \rightleftharpoons H_2$ For $P_{H_2}=1$ bar, $E_{H^+/H_2}= -0.06\,\text{pH}$ At $\text{pH}=1$, $E_{left}=-0.06\,\text{V}$

Right electrode: $AgCl(s)+e^- \rightleftharpoons Ag(s)+Cl^-$ $E_{right}=E^0-0.06\log[Cl^-]$ Given $E^0=0.22$ V.

Since the solution is HCl and at pH $=1$, $[H^+]=10^{-1}\,\text{M}$ Assuming complete dissociation of HCl, $[Cl^-]=10^{-1}\,\text{M}$ Thus, $E_{right}=0.22-0.06\log(10^{-1})$ $=0.22-0.06(-1)=0.28\,\text{V}$

Hence cell emf, $E_{cell}=E_{right}-E_{left}=0.28-(-0.06)=0.34\,\text{V}$ This is the stopping potential for sodium: $V_{s,Na}=0.34\,\text{V}$

3. Stopping potential required for potassium

$V_{s,K}=V_{s,Na}+0.05=0.34+0.05=0.39\,\text{V}$

So we need the cell emf to become $0.39$ V by changing only the pH of HCl.

4. Cell emf as a function of pH

Let pH $=x$. Then for HCl, $[H^+]=[Cl^-]=10^{-x}$

Hydrogen electrode: $E_{left}=-0.06x$

Silver-silver chloride electrode: $E_{right}=0.22-0.06\log(10^{-x})=0.22+0.06x$

Therefore, $E_{cell}=E_{right}-E_{left}=(0.22+0.06x)-(-0.06x)=0.22+0.12x$

For potassium, $0.22+0.12x=0.39$ $0.12x=0.17$ $x=\frac{0.17}{0.12}=1.4167$

So, $\text{pH} \approx 1.42$

5. Convert to the asked form

The hydrogen ion concentration is $[H^+]=10^{-1.4167}$ $=10^{-1}\times 10^{-0.4167}\approx 0.1\times 0.383\approx 0.0383$

Thus, $[HCl]\approx 3.83\times 10^{-2}$

So in the form $\underline{\hspace{1cm}}\times 10^{-2}$ the nearest integer is $4$

6. Comparison with stored answer

Stored answer is $142$, which does not match the derived result. The likely intended quantity was the pH $\times 10^{-2}$, i.e. $1.42 = 142\times 10^{-2}$ But as written, asking for the pH of aqueous HCl in the form $\_\times 10^{-2}$ corresponds to concentration, not pH. Hence the correct integer based on the wording is $4$, while $142$ corresponds to pH $=1.42$ expressed as $142\times 10^{-2}$.