$e ^{\ominus}+ AgCl _{(s)} \rightarrow Ag _{(s)}+ Cl ^{\ominus}$

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

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

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

$E =0.22+.12=34 volt$

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

$=2.3+0.34=2.64\,eV$

$\Rightarrow$ stopping potential required for $K$

$=2.64-2.25=0.39$ volt

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

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

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

$0.17=+.12 pH$

$pH =1.4166 \Rightarrow 1.42$