Problem#1
A 100-W incandescent light bulb has a cylindrical tungsten filament 30.0 cm long, 0.40 mm in diameter, and with an emissivity of 0.26. (a) What is the temperature of the filament? (b) For what wavelength does the spectral emittance of the bulb peak? (c) Incandescent light bulbs are not very efficient sources of visible light. Explain why this is so.
Answer:
(a) energy radiates at the rate H = eAσT4
The surface area of a cylinder of radius r and length l is A = 2πrl
T = (H/eAσ)1/4
T = {100 W/(2π)(0.20 x 10-3 m)(0.30 m)(0.26)(5.671 x 10-8)}1/4
T = 2.06 x 103 K
(b) λmT = 2.90 x 10-3 m.K
λm = (2.90 x 10-3 m.K)/(2.06 x 103 K) = 1410 nm
(c) λm is in the infrared. The incandescent bulb is not a very efficient source of visible light because much of the emitted radiation is ini the infrared.
Problem#2
Determine the wavelength at the peak of the Planck distribution, and the corresponding frequency ƒ, at these temperatures: (a) 3.00 K; (b) 300 K; (c) 3000 K.
Answer:
(a) λmT = 2.90 x 10-3 m.K
λm = (2.90 x 10-3 m.K)/(3.00 K) = 0.966 mm and
f = c/λm = (3.00 x 108 m/s)/( 0.966 x 10-3 m) = 3.10 x 1011 Hz
(b) A factor of 100 increase in the temperature lowers λm by a factor of 100 to 9.66 μm and raises the frekuency the same factor, to 3.10 x 1013 Hz.
(c) Similarly, λm = 966 nm and f = 3.10 x 1014 Hz
Problem#3
Radiation has been detected from space that is characteristic of an ideal radiator at (This radiation is a relic of the Big Bang at the beginning of the universe.) For this temperature, at what wavelength does the Planck distribution peak? In what part of the electromagnetic spectrum is this wavelength?
Answer:
λmT = 2.90 x 10-3 m.K
λm = (2.90 x 10-3 m.K)/(2.728 K) = 1.06 x 10-3 m = 1.06 mm
This wavelength is in the microwave portion of the electromagnetic spectrum. This radiation is often referred to as the “microwave background”.
Problem#4
The shortest visible wavelength is about 400 nm. What is the temperature of an ideal radiator whose spectral emittance peaks at this wavelength?
Answer:
λmT = 2.90 x 10-3 m.K
T = (2.90 x 10-3 m.K)/(400 x 10-9 m) = 7250 K
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