Question

4. A bulb of 40 W is producing a light of wavelength
620 nm with 80% of efficiency then the number
of photons emitted by the bulb in 20 seconds are
(1eV = 1.6 x 10-19 J, hc = 12400 eV Å)
​

Answer 1

Explanation:

ANSWER

Powerofabulb(p)=40W

energyemittedbyabulb=power×Time(s)=40×20=800J

Energyofphotonsemittedbyabulb=(

100

80

)×800=640J

Wavelength(λ)=620nm

Energyofaphoton=

λ

hc

=

620×10

−9

12400×1.6×10

−19

×10

−10

=

3.2×10

−19

640

=2×10

21

photons