Question

Nuclear fusion reactions at the center of the sun produce gamma-ray photons with energies of order 1 MeV (106 eV). By contrast, what we see emanating from the sun's surface are visible-light photons with wavelengths of order 500 nm. A simple model that explains this difference in wavelength is that a photon undergoes Compton scattering many times - in fact, about 1026 times, as suggested by models of the solar interior - as it travels from the center of the sun to its surface.

Part A

Estimate the increase in wavelength of a photon in an average Compton-scattering event.

Part B

Find the angle in degrees through which the photon is scattered in the scattering event described inpart (a). (Hint: A useful approximation is cosϕ≈1−ϕ2/2, which is valid for ϕ≪1. Note that ϕ is in radians in this expression.)

Part C

It is estimated that a photon takes about 106 years to travel from the core to the surface of the sun. Find the average distance that light can travel within the interior of the sun without being scattered. (This distance is roughly equivalent to how far you could see if you were inside the sun and could survive the extreme temperatures there. As your answer shows, the interior of the sun isvery opaque.)

Answer 1

> (a) d = hc 1026 E = 1 mew=lober, d=500x18-9 m - 4.136x1015) (29998X108) 1x1012 m 1x106 Change in wavelength (Ad) = d - d = 500x10? 181012 Adz 500X10-9m o tu compton scattering events shift is 1026 times Increase in wavelength in an average compton Scattering event is Ad = 500xlo 5x10 -33 m (b Cos d = 1-0² (5000 Ad = h (1-cosp) - he fi- (-$2)) Ad = h (62) 240 mc 2mc a h 3) (9:11 X10-31.)( 3X108) 6.42x10" rad 6.63x1534 du degrees degrees 2 mc 2(5x10-33 18= 3.7810-9 or 4X10 degrees t = 106 sec. For years a 30156 x 10'3 1026 scatterings t = 3.156X10'3 3.2 X10-13 sec 1026 t =