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

Answer:

ANSWER

Power of the bulb = 100 watt = 100Js−1

Energy of one photon is E=hν=hc/λ

where, h=6.626×10−34Js,c=3×108ms−1

Given λ=400nm=400×10−9m

By putting the values, we get

E=6.626×10−34×3×108/(400×10−9)

E=4.969×10−19J

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

n×4.969×10−19=100

n=4.969×10−19100

n=2.012×1020photons per sec

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Answer 2

Answer:

2× 10²⁰ s minus 1

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