The stopping potential

Question:

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 :

(Given : Planck's constant $(h)=6.63 \times 10^{-34} \mathrm{Js}$, electron charge $e=1.6 \times 10^{-19} \mathrm{C}$ )

  1. (1) $1.82 \mathrm{eV}$

  2. (2) $1.66 \mathrm{eV}$

  3. (3) $1.95 \mathrm{eV}$

  4. (4) $2.12 \mathrm{eV}$


Correct Option: , 2

Solution:

(2) $f_{0}=4 \times 10^{14} \mathrm{~Hz}$

$W_{0}=h f_{0}=6.63 \times 10^{-34} \times\left(4 \times 10^{14}\right) \mathrm{J}$

$=\frac{\left(6.63 \times 10^{-34}\right) \times\left(4 \times 10^{14}\right)}{1.6 \times 10^{-19}}$

$=1.66 \mathrm{eV}$

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