Question 11

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

Answer

(Given)Power of bulb, P = 25 Watt = 25 Js–1

We know, Energy of one photon, E = hν λ

Substituting the values in the given expression of E:

E = 34.87 × 10–20 J

Rate of emission of quanta per second is given by R = P/ E, Where R is the rate of emission, P is the power & E is the energy

Substituting the values in the equation we get

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