The photoelectric current from Na (work function, w0 = 2.3 eV) is stopped by the output voltage of t

Q: Why do we take the concentration of H+ and Cl- as equal?

The cell uses aqueous HCl as the electrolyte. Since HCl is a strong acid, it dissociates completely into equimolar amounts of H+ and Cl- ions.

Q: How do we know the cell potential is exactly the stopping potential?

The problem states the photoelectric current is 'stopped by the output voltage of the cell'. For the current to just stop, the applied reverse voltage must exactly equal the maximum kinetic energy of the emitted photoelectrons (in electron-volts).

Q: Why does the incident light energy remain constant?

The problem specifies 'all other conditions remaining the same' when switching from Sodium to Potassium. This implies the same light source is illuminating the new metal plate.

Q: I got 58 as the answer by following the book. Why is it wrong?

The reference book made a fundamental physics error by writing the photoelectric equation as $E = w_0 - K_{\text{max}}$ instead of $E = w_0 + K_{\text{max}}$. Kinetic energy cannot be negative! The correct application of Einstein's equation yields 142.

Q: What is the role of the Ag/AgCl electrode here?

It acts as the cathode (reduction half-cell). Its standard reduction potential ($0.22\text{ V}$) combined with the Standard Hydrogen Electrode ($0\text{ V}$) gives the cell its baseline voltage, which is then modulated by the pH.

Question

The photoelectric current from Na (work function, $w_0 = 2.3	ext{ eV}$) is stopped by the output voltage of the cell $	ext{Pt}(s) | 	ext{H}_2(g, 1	ext{ bar}) | 	ext{HCl}(aq., 	ext{pH}=1) | 	ext{AgCl}(s) | 	ext{Ag}(s)$. The pH of aqueous HCl required to stop the photoelectric current from K ($w_0 = 2.25	ext{ eV}$), all other conditions remaining the same, is ........... $	imes 10^{-2}$ (to the nearest integer).
Given, $2.303\frac{RT}{F} = 0.06	ext{ V}$; $E^\circ_{	ext{AgCl}|	ext{Ag}|	ext{Cl}^-} = 0.22	ext{ V}$

Sigma Percentile · Accepted Answer

The beauty of advanced physics and chemistry lies in their convergence. In this thrilling problem, we are tasked with using an electrochemical cell to provide the exact stopping potential required to halt photoelectrons emitted from a metal surface.

### The Grand Convergence: Physics Meets Chemistry

We have two distinct phenomena at play here. On one side, the **Photoelectric Effect** dictates that the energy of an incident photon is split into the work function of the metal and the maximum kinetic energy of the emitted electrons:
$$E_{	ext{photon}} = w_0 + K_{	ext{max}}$$

On the other side, an **Electrochemical Cell** generates a voltage based on the concentration of its electrolytes, governed by the **Nernst Equation**. The problem states that the cell's output voltage *stops* the photoelectric current. This means the cell's potential is exactly equal to the stopping potential ($V_s$):
$$K_{	ext{max}} = e V_s = e E_{	ext{cell}}$$

### Decoding the Electrochemical Engine

Let's analyze the cell: $	ext{Pt}(s) | 	ext{H}_2(g, 1	ext{ bar}) | 	ext{HCl}(aq) | 	ext{AgCl}(s) | 	ext{Ag}(s)$.

The anode is a Standard Hydrogen Electrode (SHE), where hydrogen gas oxidizes:
$$\frac{1}{2}	ext{H}_2 ightarrow 	ext{H}^+ + e^-$$

The cathode is a Silver-Silver Chloride electrode, where reduction occurs:
$$	ext{AgCl}(s) + e^- ightarrow 	ext{Ag}(s) + 	ext{Cl}^-$$

The standard cell potential is simply the cathode potential minus the anode potential:
$$E^\circ_{	ext{cell}} = 0.22	ext{ V} - 0	ext{ V} = 0.22	ext{ V}$$

Now, we apply the Nernst equation. Since $	ext{HCl}$ is a strong acid, it dissociates completely, meaning $[	ext{H}^+] = [	ext{Cl}^-]$.
$$E_{	ext{cell}} = E^\circ_{	ext{cell}} - 0.06 \log([	ext{H}^+][	ext{Cl}^-])$$
$$E_{	ext{cell}} = 0.22 - 0.06 \log([	ext{H}^+]^2)$$
$$E_{	ext{cell}} = 0.22 - 0.12 \log[	ext{H}^+]$$

Since $	ext{pH} = -\log[	ext{H}^+]$, we get a beautifully simple linear relation:
$$E_{	ext{cell}} = 0.22 + 0.12	ext{pH}$$

### The Photoelectric Bridge

For the first case using a Sodium ($	ext{Na}$) plate, the $	ext{pH}$ is given as $1$. Let's find the cell potential:
$$E_{	ext{cell, Na}} = 0.22 + 0.12(1) = 0.34	ext{ V}$$

This $0.34	ext{ V}$ acts as our stopping potential. Using Einstein's equation, we can find the energy of the incident light. Given the work function of Sodium is $2.3	ext{ eV}$:
$$E_{	ext{photon}} = 2.3	ext{ eV} + 0.34	ext{ eV} = 2.64	ext{ eV}$$

This incident light energy is a constant property of our light source and will remain unchanged.

### The Potassium Shift

Next, we replace the Sodium plate with a Potassium ($	ext{K}$) plate, which has a lower work function of $2.25	ext{ eV}$. Because the work function is lower, the electrons will be emitted with *more* kinetic energy, requiring a *higher* stopping potential.

Let's calculate the new required stopping potential:
$$2.64	ext{ eV} = 2.25	ext{ eV} + e V_{s, 	ext{K}}$$
$$V_{s, 	ext{K}} = 0.39	ext{ V}$$

Our electrochemical cell must now generate $0.39	ext{ V}$.

### The Final Calculation and The Trap

To make the cell generate $0.39	ext{ V}$, we must alter the $	ext{pH}$ of the $	ext{HCl}$ solution. We return to our simplified Nernst equation:
$$0.39 = 0.22 + 0.12	ext{pH}_{	ext{new}}$$
$$0.12	ext{pH}_{	ext{new}} = 0.17$$
$$	ext{pH}_{	ext{new}} = \frac{0.17}{0.12} = \frac{17}{12} \approx 1.416$$

Converting this to the requested format:
$$	ext{pH} = 141.6 	imes 10^{-2} \approx \mathbf{142 	imes 10^{-2}}$$

**A Crucial Warning:** The reference material for this problem contains a severe physics blunder. It incorrectly writes the photoelectric equation as $E_{	ext{photon}} = w_0 - K_{	ext{max}}$, effectively treating kinetic energy as a negative quantity. This flawed logic leads to a stopping potential of $0.29	ext{ V}$ and a final answer of $58.3$. As a student of science, you must always trust the fundamental laws of physics over a printed typo. The true, physically accurate answer is $142$.

Enter Numerical Value:

Visualized Solution

$The Grand Setup$

$Decoding the Cell$

$Nernst Equation for the Cell$

$Cell Potential for Sodium (Na)$

$Incident Photon Energy$

$Stopping Potential for Potassium (K)$

$Calculating the New pH$

$Final Answer & The Book’s Trap$

The Sigma Insight: Electrochemical Series, Electrode Potential and EMF

The Grand Convergence

Decoding the Electrochemical Engine

The Photoelectric Bridge

The Potassium Shift

The Final Calculation and The Trap

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Grand Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Decoding the Cell

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Nernst Equation for the Cell

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Cell Potential for Sodium (Na)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Incident Photon Energy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Stopping Potential for Potassium (K)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating the New pH

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer & The Book’s Trap