1e:I[80107,["/_next/static/chunks/235ad55241b6443f.js","/_next/static/chunks/8337ab1f037b6654.js","/_next/static/chunks/c9e933b5fd1a0afc.js","/_next/static/chunks/848a330faa2461ac.js","/_next/static/chunks/9f2942e368a262d2.js","/_next/static/chunks/92f9e11244990470.js","/_next/static/chunks/a8b867fbb1dfca1a.js","/_next/static/chunks/1100f0390a901fa7.js","/_next/static/chunks/bbe6bd9d3c7b0355.js","/_next/static/chunks/a82195c3e5de7dd4.js"],"default"] 1f:I[76763,["/_next/static/chunks/235ad55241b6443f.js","/_next/static/chunks/8337ab1f037b6654.js","/_next/static/chunks/c9e933b5fd1a0afc.js","/_next/static/chunks/848a330faa2461ac.js","/_next/static/chunks/9f2942e368a262d2.js","/_next/static/chunks/92f9e11244990470.js","/_next/static/chunks/a8b867fbb1dfca1a.js","/_next/static/chunks/1100f0390a901fa7.js","/_next/static/chunks/bbe6bd9d3c7b0355.js","/_next/static/chunks/a82195c3e5de7dd4.js"],"default"] 1d:T43a,{"@context":"https://schema.org","@graph":[{"@type":"BreadcrumbList","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https://www.divinejee.com"},{"@type":"ListItem","position":4,"name":"The photoelectric current from Na Work function w0 23 eV is ","item":"https://www.divinejee.com/question/the-photoelectric-current-from-na-work-function-w0-23-ev-is-stopped-by-the-output-voltage-of-the"}]},{"@type":"QAPage","mainEntity":{"@type":"Question","name":"Question The photoelectric current from Na Work function w0 23 eV is on Topic","text":"The photoelectric current from Na (Work function, w₀\n = 2.3 eV) is stopped by the output voltage of\nthe cell
Pt(s) | H₂\n(g,...","answerCount":1,"upvoteCount":0,"datePublished":"2026-01-16","author":{"@type":"Organization","name":"DivineJEE"},"acceptedAnswer":{"@type":"Answer","text":"Check the detailed solution on the page.","upvoteCount":0,"url":"https://www.divinejee.com/question/the-photoelectric-current-from-na-work-function-w0-23-ev-is-stopped-by-the-output-voltage-of-the"}}}]}20:T478,$${1 \over 2}$$H₂ + AgCl $$ \to $$ H⁺ + Ag + Cl^-

From Nernst Equation,

E_cell = $$E_{cell}^0$$ - $${{0.06} \over 1}\log \left[ {{H^ + }} \right]\left[ {C{l^ - }} \right]$$

= 0.22 – 0.06 log 10^–2 = 0.34 V

Work function of Na metal = 2.3 eV

KE of photoelectron = 0.34 eV

Energy of incident radiation = 2.3 + 0.34 = 2.64 eV

For K atom,

Energy of incident radiation for K metal = 2.64 eV

Work function of K metal = 2.25 eV

KE of photoelectrons = 2.64 – 2.25 = 0.39 eV

$$ \therefore $$ E_cell = 0.39 V

Using Nernst Equation,

0.39 = 0.22 - $${{0.06} \over 1}\log \left[ {{H^ + }} \right]\left[ {C{l^ - }} \right]$$

As $$\left[ {{H^ + }} \right] = \left[ {C{l^ - }} \right]$$

0.39 = 0.22 - $${{0.06} \over 1}\log {\left[ {{H^ + }} \right]^2}$$

$$ \Rightarrow $$ 0.39 = 0.22 - $$0.12 \times \log \left[ {{H^ + }} \right]$$

$$ \Rightarrow $$ 0.39 = 0.22 + 0.12 $$ \times $$ pH

$$ \Rightarrow $$ pH = 1.42 = 142 $$ \times $$ 10^-2