Dual Nature of Matter

JEE Main level, single correct, medium

129 of 200

Question 129 of 200

The variation of stopping potential (V0)\left({V}_{0}\right) as a function of the frequency (ν)(\nu ) of the incident light for a metal is shown in figure. The work function of the surface is

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Answer: (B) 2.07eV2.07eV

The relation between the stopping potential and the frequency of the incident light is given by
eV0=hν−ϕ...(1)e{V}_{0}=h\nu -\phi ...\left(1\right)
When V0=0{V}_{0}=0,
ϕ=hν...(2)\phi =h\nu ...\left(2\right)
Substitute the value of the frequency corresponding to zero stopping potential into equation (2) to calculate the required work function of the material.
ϕ=6.626×10−34J⋅s×5×1014Hz1.6×10−19J/eV=6.626×516≈2.07eV\begin{matrix}\phi & = & \dfrac{6.626\times {10}^{-34}J\cdot s\times 5\times {10}^{14}Hz}{1.6\times {10}^{-19}J/eV} \\ & = & \dfrac{6.626\times 5}{16}\approx 2.07eV\end{matrix}

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