The photoelectric current from Na (Work function, w = 2.3 eV) is stopped by the output voltage of th — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

**Step 1: Find the stopping potential for Na** The stopping potential equals the maximum kinetic energy of photoelectrons divided by the charge: $$V_s = \frac{KE_{max}}{e} = \frac{hν - w}{e}$$ For the given electrochemical cell, the cell potential provides the stopping voltage. At pH = 1: $$E°_{cell} = E°_{Ag/AgCl} - E°_{H⁺/H_2}$$ Using the Nernst equation for the hydrogen electrode: $$E_{H⁺/H_2} = 0 - \frac{0.059}{2} \log\frac{1}{[H⁺]} = 0.059 \text{ V (at pH = 1)}$$ The cell potential at pH = 1 stops the photoelectric current from Na: $$E_{cell} = 0.222 - 0.059 = 0.163 \text{ V (approximately)}$$ This equals the kinetic energy stopping potential for Na. **Step 2: Apply the condition for K** For K with work function 2.25 eV, using the same light source: $$KE_K = hν - 2.25 = hν - 2.3 + 0.05 = KE_{Na} + 0.05 \text{ eV}$$ The stopping potential for K: $$V_s(K) = 0.163 + 0.05 = 0.213 \text{ V}$$ **Step 3: Find required pH** $$0.213 = 0.222 - 0.059 \log\frac{1}{[H⁺]}$$ $$0.059 \log[H⁺] = 0.009$$ $$\log[H⁺] = 0.153$$ $$[H⁺] = 10^{1.847} ≈ 58.4$$ $$pH = -\log(58.4) ≈ -1.77$$ Therefore, the answer is **58.00**.