Question #179922

Our sun has a wavelength of maximum intensity of emission approximately 510 nm.


b) Using the Stefan–Boltzmann law, and given that the luminosity of the sun is 3.9 x 10^26 W, calculate the radius of the sun, giving your answer to two significant figures.


1
Expert's answer
2021-04-19T07:22:47-0400

From the Wien's displacement law:


λmax=bT\lambda_{max} =\dfrac{b}{T}

where λmax=510nm=5.1×107m\lambda_{max} = 510nm = 5.1\times 10^{-7}m is the  wavelength of maximum intensity, TT is the absolute temperature of the Sun, and b=2.9×103mKb = 2.9\times 10^{-3}m\cdot K is the Wien's displacement constant. Thus, the temperature of the Sun is:


T=bλmaxT = \dfrac{b}{\lambda_{max}}

According to the Stefan–Boltzmann law:


LA=σT4\dfrac{L}{A} = \sigma T^4

where L=3.9×1026WL = 3.9 \times 10^{26} W is the luminosity of the Sun, AA is the area of Sun's surface, σ=5.67×108W/(m2K4)\sigma = 5.67\times 10^{-8}W/(m^2\cdot K^4) is the Stefan–Boltzmann constant. Substituting the expression for TT and expressing AA, obtain:


A=LσT4=Lλmax4σb4A = \dfrac{L}{\sigma T^4} = \dfrac{L\lambda_{max}^4}{\sigma b^4}

On the other hand, the area of a sphere is:


A=πR2A = \pi R^2

where RR is the radius of the Sun. Expressing R, find:


R=Aπ=Lλmax4πσb4=λmax2b2LπσR=(5.1×107)2(2.9×103)23.9×1026π5.67×1081.45×109mR = \sqrt{\dfrac{A}{\pi}} = \sqrt{ \dfrac{L\lambda_{max}^4}{\pi\sigma b^4}}= \dfrac{\lambda_{max}^2}{b^2}\sqrt{\dfrac{L}{\pi\sigma}}\\ R = \dfrac{(5.1\times 10^{-7})^2}{(2.9\times 10^{-3})^2}\sqrt{\dfrac{3.9 \times 10^{26}}{\pi\cdot 5.67\times 10^{-8}}} \approx 1.45\times 10^{9}m

Answer. 1.45×109m1.45\times 10^{9}m


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