Q:   The outer regions of the solar atmosphere -- the photosphere --
             are much less dense than the air on Earth.  The density
             there is roughly 2.5 x 10^(-4) kg/m^3.
             Pretend that is made entirely of hydrogen atoms, each of
             which has a typical size of 10^(-10) m.

       What is the mean free path of molecules in the solar photosphere?

             The equation for mean free path of a particle is


                                                 1
              L(particle)  =  ----------------------------------------------
                               (number density) * pi * (size of particle)^2
               

             In the case of the Sun, assuming an atmosphere of pure hydrogen,
             we can compute

                                           rho         2.5 x 10^(-4) kg/m^3
                  number density  n  =  --------  =  -----------------------
                                          m(H)         1.67 x 10^(-27) kg


                                     =   1.5 x 10^(23)  atoms/m^3


                  size of particle   =    1 Angstrom  =  10^(-10) m


              and so

                                                 1
                      L(particle)  =    -------------------------------
                                        1.5 x 10^(23) * pi * 10^(-20)


                                   =    0.2 mm



       How does it compare to that of molecules in an ordinary room on Earth?


             The mfp we calculated earlier for molecules in the Earth's
             atmosphere was 

                      L(Earth air)  =   2 micrometers


             So the mfp is much larger in the solar photosphere.