Ncert Exercises & Solutions

2.11 A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57µm. Calculate the rate of emission of quanta per second.

Power of bulb, P = 25 Watt = 25 J/s
Energy of one photon, E = hν = hc/

λ

Substituting the values in the given expression of E:

E = \frac{(6.626 x 10^{- 34}) (3 x 10^{8})}{(0.57 x 10^{- 6})} = 34.87 x 10^{- 26} J

E = 34.87 ×

10^{- 20}

J
Rate of emission of quanta per second

= \frac{25}{34.87 x 10^{- 20}} = 7.169 x 10^{19} s^{- 1}