In the photosphere of the Sun, the opacity to optical light is
           very roughly

               kappa  =  0.0264  m^2 / kg

    and the density

               rho  =  2.5 x 10^-4  kg / m^3


    Q:  What is the mean free path of an optical photon in the photosphere?


                                  1
               L(photon)  =  -------------  =   1.5 x 10^5  meters
                             kappa * rho




    Q:  How does that compare to the mean free path of a hydrogen atom?


             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

 
              which is much, MUCH less than the mean free path of a photon