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The central density of the Sun is calculated to be p=1.53x10$ kg/m³ and the averaged opacity at the center to be k=0.217 m²/kg. a) Calculate the mean free path of a photon at the center of the Sun b) Calculate the average time it would take for the photon to escape from the Sun ignoring intensity changes.

The central density of the Sun is calculated to be p=1.53x10$ kg/m³ and the averaged opacity at the center to be k=0.217 m²/kg. a) Calculate the mean free path of a photon at the center of the Sun b) Calculate the average time it would take for the photon to escape from the Sun ignoring intensity changes.

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Mean free path formula is attached from the book. Thank you. 

vt FIGURE 9.6 Mean free path, l, of a hydrogen atom. atoms per cubic meter is roughly = 1.25 x 1023 m-3 where mH is the mass of a hydrogen atom. In an approximate sense, two of these 8. we may consider the equivalent problem of a single atom of radius 2ao moving with through a collection of stationary points that represent the centers of the other atom amount of time t, this atom has moved a distance vt and has swept out a cylindricalval V = 1 (2a0)²vt = o vt, where o = 1 (2a0)² is the collision cross section of the atom: classical approximation. Within this volume V are nV = nơ vt point atoms with w the moving atom has collided. Thus the average distance traveled between collision peəds %3D IGIL MIC %D no vt nơgs ne) omulov The distance l is the mean free path between collisions.° For a hydrogen atom, 10 o = 1 (2a0)² = 3.52 x 10-20 m².o deib Thus the mean free path in this situation is 00of 1. = 2.27 x 10-4 m. l = ÷ AVLK || %D ou The mean free path is several billion times smaller than the temperature scale height. As. result, the atoms in the gas see an essentially constant kinetic temperature between collisioe They are effectively confined within a limited volume of space in the photosphere. O1 this cannot be true for the photons as well, since the Sun's photosphere is the visible im 8This treats the atoms as solid spheres, a classical approximation to the quantum atom. The concept of cross section, which will be discussed in more detail in Section 10.3, actually 1ep probability of particle interactions but has units of cross-sectional area. 1ºA more careful calculation, using a Maxwellian velocity distribution for all of the atoms, resuis n path that is smaller by a factor of /2.
Transcribed Image Text:vt FIGURE 9.6 Mean free path, l, of a hydrogen atom. atoms per cubic meter is roughly = 1.25 x 1023 m-3 where mH is the mass of a hydrogen atom. In an approximate sense, two of these 8. we may consider the equivalent problem of a single atom of radius 2ao moving with through a collection of stationary points that represent the centers of the other atom amount of time t, this atom has moved a distance vt and has swept out a cylindricalval V = 1 (2a0)²vt = o vt, where o = 1 (2a0)² is the collision cross section of the atom: classical approximation. Within this volume V are nV = nơ vt point atoms with w the moving atom has collided. Thus the average distance traveled between collision peəds %3D IGIL MIC %D no vt nơgs ne) omulov The distance l is the mean free path between collisions.° For a hydrogen atom, 10 o = 1 (2a0)² = 3.52 x 10-20 m².o deib Thus the mean free path in this situation is 00of 1. = 2.27 x 10-4 m. l = ÷ AVLK || %D ou The mean free path is several billion times smaller than the temperature scale height. As. result, the atoms in the gas see an essentially constant kinetic temperature between collisioe They are effectively confined within a limited volume of space in the photosphere. O1 this cannot be true for the photons as well, since the Sun's photosphere is the visible im 8This treats the atoms as solid spheres, a classical approximation to the quantum atom. The concept of cross section, which will be discussed in more detail in Section 10.3, actually 1ep probability of particle interactions but has units of cross-sectional area. 1ºA more careful calculation, using a Maxwellian velocity distribution for all of the atoms, resuis n path that is smaller by a factor of /2.
The central density of the Sun is calculated to be p=1.53x10$ kg/m³ and the averaged opacity at the center to be k=0.217 m²/kg. a) Calculate the mean free path of a photon at the center of the Sun b) Calculate the average time it would take for the photon to escape from the Sun ignoring intensity changes.
Transcribed Image Text:The central density of the Sun is calculated to be p=1.53x10$ kg/m³ and the averaged opacity at the center to be k=0.217 m²/kg. a) Calculate the mean free path of a photon at the center of the Sun b) Calculate the average time it would take for the photon to escape from the Sun ignoring intensity changes.
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