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The work function of the nickel is $5 \mathrm{eV}$. When a light of wavelength $2000 Å$ falls on it, it emits photoelectrons in the circuit. Then the potential difference necessary to stop the fastest electrons emitted is (given $\left.h=6.67 \times 10^{-34} \mathrm{~J}-\mathrm{s}\right)$
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The correct answer is:
$1.25 \mathrm{~V}$
Stopping potential is given by
$\begin{aligned} e V_0 & =\frac{h c}{\lambda}-\phi \\ \therefore \quad e V_0 & =\frac{6.67 \times 10^{-34} \times 3 \times 10^8}{2 \times 10^{-7}}\end{aligned}$
$-5 \times 1.6 \times 10^{-19}$
$\begin{aligned} & =10.005 \times 10^{-19}-8 \times 10^{-19} \\ & =2.005 \times 10^{-19}\end{aligned}$
or $\quad \begin{aligned} V_0 & =\frac{2.005 \times 10^{-19}}{1.6 \times 10^{-19}} \text { volt } \\ & =1.25 \mathrm{~V}\end{aligned}$
$\begin{aligned} e V_0 & =\frac{h c}{\lambda}-\phi \\ \therefore \quad e V_0 & =\frac{6.67 \times 10^{-34} \times 3 \times 10^8}{2 \times 10^{-7}}\end{aligned}$
$-5 \times 1.6 \times 10^{-19}$
$\begin{aligned} & =10.005 \times 10^{-19}-8 \times 10^{-19} \\ & =2.005 \times 10^{-19}\end{aligned}$
or $\quad \begin{aligned} V_0 & =\frac{2.005 \times 10^{-19}}{1.6 \times 10^{-19}} \text { volt } \\ & =1.25 \mathrm{~V}\end{aligned}$
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