Physics

Problem

A tungsten filament in an electric bulb may be treated as a blackbody radiator. When operating at its rated conditions, the bulb consumes

60\,\text{W}

of electrical power, all of which is assumed to appear as thermal radiation. At this condition the peak of its emission spectrum occurs at

\lambda_{\text{max}} = 970\,\text{nm}

. Use

b = 2.9\times10^{-3}\,\text{m\,K}

and

\sigma = 5.67\times10^{-8}\,\text{W\,m}^{-2}\,\text{K}^{-4}

. a. (i) Determine the operating temperature of the filament. b. (ii) Estimate the effective radiating surface area of the filament. c. (iii) Over time, the filament thins and its resistance increases so that, when connected to the same supply voltage, the electrical power drawn falls to

45\,\text{W}

. Assuming the surface area remains the same and all the power is still emitted as thermal radiation, calculate the new wavelength corresponding to maximum intensity.

Intuitive explanation

The key is recognizing that a blackbody's peak wavelength and total radiated power are governed by two separate laws: Wien's displacement law tells you the filament's temperature from $\lambda_{\text{max}}$ , while Stefan–Boltzmann law connects that temperature to the surface area needed to radiate 60 W. You'll find the temperature first, then use it to solve for the filament's radiating surface area.

Intuitive explanation

Browse the problem library