Question

A 100 watt bulb emits monochromatic light of wavelength 400 nm. Calculate the number of photons emitted per second by the bulb.

Answer 1

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Answer 2

Answer:

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Power of the bulb = 100 watt = 100J${s}^{ - 1}$

Energy of one photon is E=hν=hc/λ

where, h=6.626×${10}^{ - 34}$Js,c=3×${10}^{8}$m [tex] {s}^{ - 1} [/tex]

Given λ=400nm=400×${10}^{ - 9}$m

By putting the values, we get

$</em><em>E</em><em> = 6.626 \times {10}^{ - 34} \times 3 \times {10}^{8} </em><em>/</em><em> </em><em>(400 \times {10}^{ - 9} )$

E=4.969×${10}^{ - 19}$J

Number of photons emitted in 1 sec × energy of one photon = power

n×4.969×${10}^{ - 19}$=100

$n = \frac{100}{4.969 \times {10}^{ - 19} }$

$n = 2.012 \times {10}^{20}$photons per sec.

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