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JEE Mains · Physics · STD 12 - 11. Dual nature of radiation and matter
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 \(e = 1.6 \times 10^{-19}\, C\))
- A \(1.82\)
- B \(1.66\)
- C \(2.12\)
- D \(1.95\)
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 \(v=4 \times 10^{14} \mathrm{Hz}\) \(0=\frac{h v}{e}-\frac{\phi}{e}\) \(\Rightarrow \phi=\mathrm{hv}\)…
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