The number of photons of wavelength 540 nm emitted per second by an electric bulb of power 100 W is (taking $h = 6 \times 10^{- 34} J - \sec$ )

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100

1000

$3 \times 10^{20}$

$3 \times 10^{18}$

answer is C.

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Detailed Solution

$p = \frac{nhc}{λt} \Rightarrow 100 = \frac{n \times 6 \times 10^{- 34} \times 3 \times 10^{8}}{540 \times 10^{- 9} \times 1} \Rightarrow n = 3 \times 10^{20}$

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