If the filament of a 100W bulb has an area 0. 25 cm ^ 2 and behaves...

In the bulb filament given, energy radiated per second per m^{2} of its surface area is given as

$\text{E} = \dfrac{P}{A} = \dfrac{100}{\text{0.25} \times 1{0}^{- 4}} = 4 \times 1{0}^{6} J s^{-1}{ m}^{- 2}$

If T is the temperature of the filament then according to Stefan's law, we have

$\text{E} = \sigma T^{4}$

or $4 \times 1 {0}^{6} = \text{5.67} \times 1 {0}^{- 8} \times T^{4}$

or $T^{4} = \dfrac{4 \times 1 {0}^{6}}{\text{5.67} \times 1 {0}^{- 8}} = \text{7.055} \times 1 {0}^{1 3}$

or $\text{T} = {\left[\text{7.055} \times 1 {0}^{1 3}\right]}^{1 / 4} = \text{2898.14 K}$

If the filament radiates the maximum energy at a wavelength ${λ}_{m}$, from Wein's displacement law, we have

${λ}_{m} T = \text{b}$

or ${λ}_{m} = \dfrac{\text{b}}{\text{T}}$

$= \dfrac{\text{2.89} \times 1{0}^{- 3}}{2898.14} = 9971.9 Å$