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 form `K(w_(0) = 2.25 eV)`, all other conditions remaining the same, is _______ `xx10^(-2)` (to the nearest integer).
Given,
`2.303(RT)/(F)=0.06V,E_(AgCl|Ag|Cl^(-))^(@)=0.22V`

Explanation

To solve the problem, we need to find the pH of the HCl solution required to stop the photoelectric current from potassium (K) with a work function of 2.25 eV, given that the photoelectric current from sodium (Na) with a work function of 2.3 eV is stopped by the output voltage of the cell.

### Step-by-Step Solution:

1. **Identify the energy of the photon for Na:**
\[ E_{\text{photon, Na}} = W_0 = 2.3 \text{ eV} \]

2. **Calculate the cell potential for Na:**
The cell potential can be calculated using the Nernst equation:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{2.303 \cdot RT}{F} \log [H^+] \]
Given \( E^\circ_{\text{cell}} = 0.22 \text{ V} \) and \( \frac{2.303 \cdot RT}{F} = 0.06 \text{ V} \) at pH = 1 (where \( [H^+] = 10^{-1} \)):
\[ E_{\text{cell, Na}} = 0.22 - 0.06 \cdot \log(10^{-1}) = 0.22 - 0.06 \cdot (-1) = 0.22 + 0.06 = 0.28 \text{ V} \]

3. **Calculate the energy of the photon for K:**
\[ E_{\text{photon, K}} = W_0 = 2.25 \text{ eV} \]

4. **Set up the equation for K:**
The energy of the photon for K must equal the work function minus the cell potential:
\[ E_{\text{photon, K}} = W_0 - E_{\text{cell, K}} \]
Rearranging gives:
\[ E_{\text{cell, K}} = W_0 - E_{\text{photon, K}} = 2.25 - E_{\text{cell, K}} \]

5. **Equate the two cell potentials:**
Since the conditions remain the same, we can equate the two cell potentials:
\[ E_{\text{cell, K}} = E_{\text{cell, Na}} + (2.3 - 2.25) = 0.28 + 0.05 = 0.33 \text{ V} \]

6. **Use the Nernst equation for K:**
Substitute \( E_{\text{cell, K}} \) back into the Nernst equation:
\[ 0.33 = 0.22 - 0.06 \log [H^+] \]

7. **Solve for \([H^+]\):**
Rearranging gives:
\[ 0.33 - 0.22 = -0.06 \log [H^+] \]
\[ 0.11 = -0.06 \log [H^+] \]
\[ \log [H^+] = -\frac{0.11}{0.06} \approx -1.8333 \]
\[ [H^+] \approx 10^{-1.8333} \approx 0.0147 \text{ M} \]

8. **Calculate the pH:**
\[ pH = -\log [H^+] \approx -\log(0.0147) \approx 1.83 \]

9. **Express the pH in the required format:**
\[ pH \approx 1.83 \approx 18.3 \times 10^{-2} \]
Rounding to the nearest integer gives:
\[ \text{Final answer: } 18 \]