# The photoelectric current from $mathrm{Na}$

Question:

The photoelectric current from Na (work function, $\mathrm{w}_{0}=2.3 \mathrm{eV}$ ) is stopped by the output voltage of the cell

$\mathrm{Pt}(\mathrm{s})\left|\mathrm{H}_{2}(\mathrm{~g}, 1 \mathrm{bar})\right| \mathrm{HCl}(\mathrm{aq} ., \mathrm{pH}=1)|\mathrm{AgCl}(\mathrm{s})| \mathrm{Ag}(\mathrm{s})$

The $\mathrm{pH}$ of aq. $\mathrm{HCl}$ required to stop the photoelectric current from $\mathrm{K}\left(\mathrm{w}_{0}=2.25 \mathrm{eV}\right)$, all other conditions remaining the same, is

$\times 10^{-2}$ (to the nearest integer).

Given, $2.303 \frac{\mathrm{RT}}{\mathrm{F}}=0.06 \mathrm{~V} ; \mathrm{E}_{\mathrm{AgCl}|\mathrm{Ag}| \mathrm{Cl}^{-}}^{0}=0.22 \mathrm{~V}$

Solution:

$\frac{1}{2} \mathrm{H}_{2} \rightarrow \mathrm{H}^{+}+\mathrm{e}^{\Theta}$

$\frac{\mathrm{e}^{\Theta}+\mathrm{AgCl}_{(\mathrm{s})} \rightarrow \mathrm{Ag}_{(\mathrm{s})}+\mathrm{Cl}^{\Theta}}{\frac{1}{2} \mathrm{H}_{2}+\mathrm{AgCl}_{(\mathrm{s})} \rightarrow \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{Ag}_{(\mathrm{s})}+\mathrm{Cl}_{(\mathrm{aq})}^{\ominus}}$

$\mathrm{E}=\varepsilon^{0}-\frac{.06}{1} \log \frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{Cl}^{\Theta}\right]}{\mathrm{P}_{\mathrm{H}_{2}}^{\frac{1}{2}}}$

$\mathrm{E}=0.22-.06 \log \frac{\left(10^{-1}\right)\left(10^{-1}\right)}{1^{\frac{1}{2}}}$

$\mathrm{E}=0.22+.12=.34$ volt

$\Rightarrow$ total energy of photon will be (for $\mathrm{Na}$ )

$=2.3+0.34=2.64 \mathrm{eV}$

$\Rightarrow$ stopping potential required for $\mathrm{K}$

$=2.64-2.25=0.39$ volt

$\mathrm{E}=\varepsilon^{0}-\frac{.06}{1} \log \frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{Cl}^{-}\right]}{\mathrm{P}_{\mathrm{H}_{2}}^{\frac{1}{2}}}$

as $\left[\mathrm{H}^{+}\right]=\left[\mathrm{Cl}^{\ominus}\right]$ so

$0.39=0.22-.06 \log \frac{\left[\mathrm{H}^{+}\right]^{2}}{1^{\frac{1}{2}}}$

$0.17=+.12 \mathrm{pH}$

$\mathrm{pH}=1.4166 \Rightarrow 1.42$