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Q.

A bulb of 40 W is producing a light of wavelength 620 run with 80% of efficiency, then the number of photons emitted by the bulb in 20 seconds are :
$(1 eV = 1.6 \times 10^{- 19} J, h c = 12400 eV)$

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a

$10^{21}$

b

$2 \times 10^{18}$

c

$10^{18}$

d

$2 \times 10^{21}$

answer is D.

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Detailed Solution

Number of photons emitted by the bulb in 20 seconds are

$[\frac{40 \times 0.8}{1.6 \times 10^{- 19}}] \times 20 = N \times \frac{1240}{620} \Rightarrow N = 2 \times 10^{21}$

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A bulb of 40 W is producing a light of wavelength 620 run with 80% of efficiency, then the number of photons emitted by the bulb in 20 seconds are :1eV=1.6×10−19J, hc=12400eV