3. (3 Points) Here, we show the distribution of solar emittance as function of p

Question

3. (3 Points) Here, we show the distribution of solar emittance as function of photon energy E Max Planck could derive its functional form with the Quantum-Hypothesis: Each body radiates by throwing quantised portions of energy in all directions, quanta of light, massless particles called photons with energy, related to a frequency v by E hv. The higher its temperature T the more generously it radiates energy, h is Planck's constant h = 6.626 × 10-34Js 0.5 0 4 0,00- '«lev 3 E=hv Estimate the temperature of the sun T, from the distribution of the emittance as function of energy, using the appropriate form of Wien's law. Use = 1.38 x 10-23J/K and hc/kB = 015Km Derive the Stefan-Boltzmann law from the Planck distribution for the emittance. (1 Bonus point: derive the numerical value of the Stefan-Boltzmann constant by evaluating the integral numerically)

Explanation / Answer

a. From weins displacement law

maximum wavelength emitted by a black body is related to its temperature as under

lambda-max = b/T

where b = 2.897*10^ -3 m K

now, form the plot, Emax = 1.5 eV

so hc/lambda-max = 1.5*1.6*10^-19 J

lambda-max = 6.63*10^-34*3*10^8/1.5*1.6*10^-19 = 828.75*10^-9 m

hence T = 3495.6259 K

b. from planks radiation formula

the radiated power per unit area as a funciton of wavelength is

(dP/d(lambda)/A = 2*pi*hc^2/(lambda^5(e^(hc/lambda*(kT)) - 1))

so the integrated power per unit area is

P/A = 2*pi*hc^2 [integrate from 0 to infinity] dlambda/(lambda^5*((e^(hc/lambda*(kT)) - 1))))

put x = hc/lambda*kT

dx = -hc*d*lambda/lambda^2kT

P/A = 2*pi*(kT)^4/h^3c^2 [integrate] x^3 dx / (e^x - 1)

integrating

P/A = 2*pi^5 k^4 T^4 / 15h^3c^2 = sigmaT^4

where sigma = 2*pi^5*k^4/15h^3c^2 = 5.67*10^-8

Related Questions

Navigate

• Browse (All)
• Browse 3
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.