Chapter 13: Problem 78
Three discs \(A, B\) and \(C\) having radii \(2 \mathrm{~m}, 4 \mathrm{~m}\) and \(6 \mathrm{~m}\) respectively are coated with carbon black on their other surfaces. The wavelengths corresponding to maximum intensity are \(300 \mathrm{~nm}, 400 \mathrm{~nm}\) and \(500 \mathrm{~nm}\) respectively. The power radiated by them are \(Q_{a}, Q_{b}\) and \(Q_{c}\) respectively (a) \(Q_{a}\) is maximum (b) \(Q_{b}\) is maximum (c) \(Q_{c}\) is maximum (d) \(Q_{a}=Q_{b}=Q_{c}\)
Short Answer
Step by step solution
Identify the given quantities
Apply Wien's Displacement Law
Calculate Temperatures
Use Stefan-Boltzmann Law to Determine Power Radiated
Calculate Areas of Discs
Compute Radiated Powers
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Wien's Displacement Law
- As an object's temperature increases, the peak of its emitted radiation shifts to shorter wavelengths.
- This law allows us to determine the temperature of an object by observing the peak wavelength of its emitted radiation.
Blackbody Radiation
- All bodies emit radiation, but a blackbody is a perfect model that absorbs all incident radiation without reflecting any.
- The amount and type of radiation emitted depends on the body's temperature.
Surface Area Calculation
- A larger surface area means more space over which radiation can be emitted.
- In scenarios involving blackbody radiation, this becomes a crucial factor in determining total emitted energy.