PLEASE ONLY USE THE INFORMATION PROVIDED. The Luminosity of the Sun from Random

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PLEASE ONLY USE THE INFORMATION PROVIDED.

The Luminosity of the Sun from Random Walking: The conditions inside the Sun imply that a photon has a mean free path length of about l = 0.5 cm; that means on average a photon "flies" about half of a centimetre before interacting or bumping into matter. Imagine a photon emitted at the center of the Sun, but only in one dimension-it can move right (+0.5 cm) or left (-0.5 cm) at each time step with equal probability. The average position is zero, the photon has no net drift pushing it to the left or to right. However, the square of the displacements does have a net drift-even though on average photons are just as likely to be to the left or to the right, over a long period of time they will have had a chance to wander a long way from the centre, some far to the left and some far to the right (and no, they are not political photons!). The mean square distance after N steps is R^2 = Nl^2. In three dimensions, the result becomes 3R^2 = Nl^2. Therefore, the number of steps that a photon needs to take on average to get to the photosphere from the center is N = 3R_s^2|l^2. The random walk time, t, is just the total distance covered divided by the speed of light (c = 3.0 times 10^10 cm/s), t = Nl/c = 3R_s^2/(lc). A hot gas at temperature T (in Kelvin) has radiation energy per unit volume of aT^4, where a = 7.56 times 10^-15 erg cm^-3 K^-4. The "leakage" of photons from the Sun-which is the Sun's luminosity-must be: L_s = (volume) times (radiation per unit volume)/random walk time to cover distance R_5 = (4 pi R_S^3/3)(aT^4)/3R_S^2/(lc) (a) Using the Sun's core temperature that you calculated in the previous question, what does the above argument give for the Suns luminosity. Compare your result with the observed value of L_S= 3.90 times 10^33 erg/sec. How does the comparison work when using a temperature of T = 4.5 times 10^6 K? What does this result tell you about the average temperature of the Sun compared to the core temperature?

Explanation / Answer

Now, From the previous answer
T = 17428831.58 K at the centre of the sun

a) Ls = (4*pi*Rs) (aT^4)*lc / 9
   here
   Rs = 6.96*10^10 cm
   a = 7.56*10^-15 erg cm^-3 /K^4
   l = 0.5 cm
   c = speed of light = 3*10^10 cm/s
   Ls = 9.147*10^34 erg/s

here Ls given = 3.9*10^33 erg/s
Ls / Ls given = 23.454

usint T = 4.5*10^6 K
Ls = 4.064*10^32 erg/s
Ls/Ls new = 225.020

hence, Core temperature is very high when compared to the average temperature of the sun

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