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QJEEM23Ph757f: JEE Main 2023 Physics - Class 12 JEEM23Ph757f Solution: The variation of stopping potential as a function of the…

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

JEE Main 2023 (Online) 10th April Evening Shift Physics - Dual Nature of Radiation Question 42 English

Multiple Choice Options

  • A 1.36 eV
  • B 18.6 eV
  • C 2.98 eV
  • D 2.07 eV

Step-by-Step Verified Solution

amp;eV0=hcλ−ϕamp;V0=hceλ−(ϕe)=hve−(ϕe)\begin{aligned}& e V_0=\frac{h c}{\lambda}-\phi \\\\& V_0=\frac{h c}{e \lambda}-\left(\frac{\phi}{e}\right)=\frac{h v}{e}-\left(\frac{\phi}{e}\right)\end{aligned}

When V0=0V_0=0 ,

ϕeamp;=(hve)ϕamp;=hv=6.626×10−34×5×10141.6×10−19amp;=6.626×51.6×10−1amp;=6.626×516≈2.07eV\begin{aligned}\frac{\phi}{e} & =\left(\frac{h v}{e}\right) \\\\\phi & =h v=\frac{6.626 \times 10^{-34} \times 5 \times 10^{14}}{1.6 \times 10^{-19}} \\\\& =\frac{6.626 \times 5}{1.6} \times 10^{-1} \\\\& =\frac{6.626 \times 5}{16} \approx 2.07 \mathrm{eV}\end{aligned}

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