Question
Light source having wavelength 331 nm is used to generate photo-electrons whose stopping potential is 0.2 V . The work function of the used metal in the experiment is $\alpha \times 10^{- 19}\text{ }J$. The value of $\alpha$ is
$\_\_\_\_$ .
( $h = 6.62 \times 10^{- 34}\text{ }J\text{ }s,e = 1.6 \times 10^{- 19}C$ and c $= 3 \times 10^{8}\text{ }m/s$ )
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Solution
Sol. [Key Concept: Photoelectric equation: $\varphi = \frac{hc}{\lambda} - {eV}_{s}$.]
$${\ V_{s} = \frac{1}{e}\left( \frac{hc}{\lambda} - \phi \right) }{\phi = \frac{hc}{\lambda} - {ev}_{s} }{= \left( \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8}}{331 \times 10^{- 9}} \right) - \left( 1.6 \times 10^{- 19} \times 0.2 \right) }{= 5.68 \times 10^{- 19}\text{ }J }$$
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