If the filament of a 100 W bulb has an area 0 . 25 cm 2 and behaves as a perfect black body, find the wavelength corresponding to the maximum in its energy distribution. [Given σ = 5.67 × 1 0 - 8 J m - 2 s K - 4 , b = 2 . 89 × 10 - 3 m K ]

Step-by-step solution

In the bulb filament given, energy radiated per second per m 2 of its surface area is given as E = P A = 1 0 0 0.25 × 1 0 - 4 = 4 × 1 0 6 J s -1 m - 2 If T is the temperature of the filament then according to Stefan's law, we have E = σ T 4 or 4 × 1 0 6 = 5.67 × 1 0 - 8 × T 4 or T 4 = 4 × 1 0 6 5.67 × 1 0 - 8 = 7.055 × 1 0 1 3 or T = 7.055 × 1 0 1 3 1 / 4 = 2898.14 K If the filament radiates the maximum energy at a wavelength λ m , from Wein's displacement law, we have λ m T = b or λ m = b T = 2.89 × 1 0 - 3 2898.1