Question

Calculati
monochromatic
500 nm
Bulb
of photos
emitted by 60 watt
bulb emitting light of wavelength
efficiency
50%

Answer 1

Explanation:

Power of the bulb = 100 watt = 100Js

−1

Energy of one photon is E=hν=hc/λ

where, h=6.626×10

−34

Js,c=3×10

8

ms

−1

Given λ=400nm=400×10

−9

m

By putting the values, we get

E=6.626×10

−34

×3×10

8

/(400×10

−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=

4.969×10

−19

100

n=2.012×10

20

photons per sec