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The variation of stopping potential (V_{0}) as a function of the frequency(v) of the incident light for a metal is
shown in figure. The work function of the surface is

 

Option: 1

2.07\; eV


Option: 2

18.6\: eV


Option: 3

2.98\: eV


Option: 4

1.36\: eV


Answers (1)

Work function (\phi)=h v^{\text {th }}

\begin{aligned} & =6.6 \times 10^{-34} \times 5 \times 10^{14} \\ & =33 \times 10^{-20} \\ & \phi=3.3 \times 10^{-19} \mathrm{~J} \\ & =\frac{3.3 \times 10^{-19}}{1.6 \times 10^{-19}} \mathrm{ev} \Rightarrow 2.07 \end{aligned}

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Ramraj Saini

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