In the photoelectric experiment, if we use a monochromatic light, the I-V curve is as shown. If work function of the metal is 2 eV, estimate the power of light used (in W). Assume efficiency of photo emission to be $10^{- 3} %$ , i.e, number of photoelectrons emitted are $10^{- 3} %$ of number of photons incident on metal. $(1.0 eV = 1.6 \times 10^{- 19} J)$

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Detailed Solution

$e V_{s} = h v - W \Rightarrow h v = 7 e V$

So, total number of photons incident is
$n = \frac{P}{h v}$
Since, $η = 10^{- 3} %$
$\Rightarrow n_{e m i t t e d} = n \times \frac{10^{- 3}}{100} \Rightarrow i = \frac{q}{t} = (\frac{n_{e m i t t e d}}{t}) e$

Where, from the graph
$i = 10 \times 10^{- 6} A \Rightarrow 10 \times 10^{- 6} = \frac{P}{h v} \times \frac{10^{- 3}}{100} \times 1.6 \times 10^{- 19} \Rightarrow 10 \times 10^{- 6} = \frac{P \times 10^{- 3} \times 1.6 \times 10^{- 19}}{7 \times 1.6 \times 10^{- 19} \times 100} \Rightarrow P = 7 W$

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