(1) pi

(2)

(a)
    Solar Mass = 1.99 × 10³⁰ kg.
    
    Solar Radius = 6.96 × 10¹⁰ cm.
    
    Solar Volume = 4/3 π R³
                 = 1.410 × 10³³ cm³.
    
    H⁺ mass = 1.67 × 10²⁷ kg.
    
    H⁺ density = Solar Mass/Solar Volume/H⁺ Mass
               = 8.435 × 10²³ cm⁻³.
    
    e⁻ density = H⁺ density.

(b)
    mean free path = l = κ⁻¹.

    κ = n[e⁻] × σ[thompson]
      = 0.506 cm⁻¹.

    l = 1.98 cm.

(c)
    For average κ,
        τ = κ × Solar Radius
          = 3.52 × 10¹⁰.

(d)
    AGN3 gives an approximation for luminosity from bremsstrahlung as, assuming H⁺ is the only ion, in erg cm⁻³ s⁻¹,
        L[ff] = 1.42 × 10⁻²⁷ Z² T⁰˙⁵ g[ff] nₑ H⁺.

    The gaunt correction factor
        g[ff] = 1.3.

    T = 106 K.

    L[ff] = 1.42 × 10⁻²⁷
          = 1.35 × 10²² erg cm⁻³ s⁻¹.

    This is the luminosity from Brems. radiation per unit volume, but has to be integrated while considering the optical depth to find the total emission from Brems. radiation. We will assume the sun is optically thin, and by our derivation in the lecture, the optical depth then is no longer important.

    The luminosity from the emissivity is, for optically thin conditions and constant emissivity j = L[ff]/4π, 
        L = ∫ 4 π j dV 
          = L[ff] × V.

    Note L[ff] includes the solid angle 4π across which each unit volume emits.

    Integrating over the sun's volume, then

    L[total,ff] = L[ff] × V
                = 1.91 × 10⁵⁵ ergs s⁻¹.

(e)

    This number is way too huge. The total luminosity of the sun is on the order of 10³³ erg s⁻¹, 22 orders of magnitude smaller. I think this means I have a conceptual problem, somewhere. I don't think optically thin is a good assumption for the sun.