Question

A 150 W bulb radiates monochromatic light of wavelength 800nm number of photons emitted per second is​

Answer 1

Answer:

6.04 x 10^20 PHOTONS PER SEC

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^ /(800×10^−9 )

E=2.48×10  ^−19  J

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

n×4.969×10  ^−19 =150  

n=  150/2.48×10  ^−19    

n=6.04×10  ^20  photons per sec