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JEE Mains · Physics · STD 12 - Dual nature of radiation and matter
  • JEE Mains 2019
  • Medium
  • MCQ

The stopping potential \(V_0\) (in \(volt\)) as a function of frequency \((v)\) for a sodium emitter, is shown in the figure. The work function of sodium, from the data plotted in the figure, will be : ................. \(eV\) (Given : Planck's constant \((h) = 6.63 \times 10^{-34}\, Js\), electron charge \(e = 1.6 \times 10^{-19}\, C\))

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Answer & Solution

Correct Answer

(B) \(1.66\)

Step-by-step Solution

Detailed explanation

\(\mathrm{hv}=\phi+\mathrm{ev}_{0}\)

\(v_{0}=\frac{h v}{e}-\frac{\phi}{e}\)

\(\mathrm{v}_{0}\) is zero for \(v=4 \times 10^{14} \mathrm{Hz}\)

\(0=\frac{h v}{e}-\frac{\phi}{e}\)

\(\Rightarrow \phi=\mathrm{hv}\)

\( = \frac{{6.63 \times {{10}^{ - 34}} \times 4 \times {{10}^{14}}}}{{1.6 \times {{10}^{ - 19}}}}\) \( = 1.66\,\,{\text{eV}}\)

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