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Question

A 100 watt bulb emits monochromatic light of wavelength 400 nm. Calculate the number of photons emitted per second by the bulb.
  1. 2×1020s1
  2. 3×1020s1
  3. 2×1020s1
  4. 1×1020s1

A
2×1020s1
B
3×1020s1
C
2×1020s1
D
1×1020s1
Solution
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Power of the bulb = 100 watt = 100Js1

Energy of one photon is E=hν=hc/λ

where, h=6.626×1034Js,c=3×108ms1

Given λ=400nm=400×109m

By putting the values, we get

E=6.626×1034×3×108/(400×109)

E=4.969×1019J

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

n×4.969×1019=100

n=1004.969×1019

n=2.012×1020photons per sec

So, option C is correct

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