The photoelectric current from Na (work function, . w _ 0 =2.3 , … - Vidyadip

MCQ

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

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

The $pH$ of aq. HCl required to stop the photoelectric current from $K \left( w _{0}=2.25 eV \right),$ all other conditions remaining the same, is..........$\times 10^{-2}$ (to the nearest integer).

Given, $2.303 \frac{ RT }{ F }=0.06 V ; E _{ AgC1|Ag|C ^{-}}^{0}=0.22\, V$

A
$146$
B
$150$
✓
$142$
D
$154$

✓

Answer

Correct option: C.

$142$

c
$\frac{1}{2} H _{2} \rightarrow H ^{+}+ e ^{\ominus}$

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

_______________________________________________________

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

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

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

$E =0.22+.12=34 volt$

$\Rightarrow$ 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

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

$0.17=+.12 pH$

$pH =1.4166 \Rightarrow 1.42$

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