Question

Find out the number of photons emitted by a 60 watt light bulb per second if 10% of electrical energy is supplied to an incandescent light bulb is radiated as visible light of wave length 6000a is

Answer 1

The number of photons emitted is n = 1.8 x 10^19

Explanation:

10 /100 x 60 J = n h c / wavelength

Plank's constant "h" = 6.63 x 10^-34

Speed of light "c" = 3 x 10^8 m/s

Wavelength = 6000 x 10^-10 m

6 = n x 6.63 x 10^-3 x 10^8/6000 x 10^-10

n = 1.8 x 10^19

Hence the number of photons emitted is n = 1.8 x 10^19

Also learn more

A photon of wavelength 4 × 10–7 m strikes on metal surface, the work function of the metal being 2.13 eV. Calculate

(i) the energy of the photon (eV),

(ii) the kinetic energy of the emission, and

(iii) the velocity of the photoelectron (1 eV= 1.6020 × 10–19 J).

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