Apparent brightness of Eta Cassiopeiae A = 1.1 × 10–9 Wm–2Apparent brightness of Eta Cassiopeiae B = 5.4 × 10–11 Wm–2Luminosity of the Sun, L = 3.8 × 1026 W(i) The peak wavelength of radiation from Eta Cassiopeiae A is 490 nm. Show thatthe surface temperature of Eta Cassiopeiae A is about 6000 K
Question
Apparent brightness of Eta Cassiopeiae A = 1.1 × 10–9 Wm–2Apparent brightness of Eta Cassiopeiae B = 5.4 × 10–11 Wm–2Luminosity of the Sun, L = 3.8 × 1026 W(i) The peak wavelength of radiation from Eta Cassiopeiae A is 490 nm. Show thatthe surface temperature of Eta Cassiopeiae A is about 6000 K
Solution
To find the surface temperature of Eta Cassiopeiae A, we can use Wien's Law, which states that the peak wavelength of radiation from a black body is inversely proportional to its temperature. The formula for Wien's Law is:
λ_max = b/T
where λ_max is the peak wavelength, T is the temperature, and b is Wien's constant (approximately 2.898 x 10^-3 m.K).
Given that the peak wavelength of radiation from Eta Cassiopeiae A is 490 nm (or 490 x 10^-9 m), we can rearrange Wien's Law to solve for T:
T = b/λ_max
Substituting the given values:
T = (2.898 x 10^-3 m.K) / (490 x 10^-9 m)
T = 5918.37 K
This value is approximately 6000 K, so the surface temperature of Eta Cassiopeiae A is about 6000 K.
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