The stopping potential \(V_0\) (in volt) as a function of frequency (\(\upsilon\)) for a sodium emitter, is shown in the figure. The work function of sodium, from the data plotted in the figure, will be:
(Given: Planck's constant (h) = \(6.63 \times 10^{-34}\) Js, electron charges e = \(1.6 \times 10^{-19}\) C)
(Given: Planck's constant (h) = \(6.63 \times 10^{-34}\) Js, electron charges e = \(1.6 \times 10^{-19}\) C)
Show Hint
- 1.95 eV
- 2.12 eV
- 1.82 eV
- 1.66 eV
The Correct Option is D
Solution and Explanation
In the graph the line meets the frequency axis at \(\upsilon_0 = 4\times10^{14}\) Hz.
\(\phi = h\upsilon_0 = 6.63\times10^{-34}\times4\times10^{14} = 2.652\times10^{-19}\) J.
In eV: \(\phi = \dfrac{2.652\times10^{-19}}{1.6\times10^{-19}} = 1.6575 \approx 1.66\) eV.
Answer: (4) 1.66 eV
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