A bulb of 40W is producing a light of $λ$ 620nm with 80% of efficiency then the number of photons emitted by the bulb in 20 sec [1ev = 1.6×10^–19J, hc=12400ev A⁰]

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2 × 10¹⁸

10²¹

2 × 10²¹

10¹⁸

answer is D.

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Detailed Solution

$p o w e r = \frac{n h c}{λ t} = 40 \times \frac{80}{100} = \frac{n \times 6.62 \times 10^{- 34} \times 3 \times 10^{8}}{620 \times 10^{- 9} \times 20} n = 2 \times 10^{21}$

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