as-500/hw2/text

(1)

Solar Luminosity, according to this page, is 3.828 × 10²⁶ W.

The conversion from W to erg/s is (1 W)/(10⁷ erg/s).

3.828 × 10²⁶ W × (10⁷ erg/s) / (1 W) = 3.828 × 10³³ erg/s.

(2)

It quotes the most powerful quasars exceeding 10⁴¹ W.

10⁴¹ W × (10⁷ erg/s) / (1 W) = 10⁴⁸ erg/s.

The article claims that 3C 273 is about 4 trillion times as bright as our sun (intrinsically).

3.828 × 10³³ erg/s × 4 × 10¹² = 1.531 × 10⁴⁶ erg/s.

(3)

Flux F observed by some distant observer is related to intrinsic luminosity L by the inverse square law and the encompassing solid angle 4 π. We use 1 AU as the distance light travels from the sun to the top of earth's atmosphere, but of course in reality the light travels from the sun's photosphere (essentially the outer edge of the sun) and strikes the earth's atmosphere at essentially the edge of the earth's sphere. To compute this more accurately, we could subtract the radii of the earth and sun from the distance. Again, we'll just use 1 AU for these calculations, as indicated by the problem.

F = L/(4 π d²) = (3.828 × 10³³ erg/s) / ( 12.566 × (1 AU)² )

F   = (3.046 × 10³² erg/s) / (1 AU)²
    = (3.046 × 10³² erg/s) / (1.4960 × 10¹³ cm)²
    = (3.046 × 10³² erg/s) / 3.3480 × 10²⁶ 
    = 9.098 × 1⁵ erg/s/cm²

(4)

I note some useful conversions and constants here, then convert the wavelength to cm. To convert from wavelength λ to energy E or frequency ν, I use the DeBroglie relationships and the definition of wave velocity. I also note that one Rydberg is 13.606 eV, the energy needed to ionize hydrogen with an electron in the ground state.

1 Å = 10⁻⁸ cm.
h = 4.13567 × 10⁻¹⁵ eV s.
c = 29979245800 cm / s.

λ   = 5000 Å × (10⁻⁸ cm/1 Å)
    = 5 × 10⁻⁵ cm.

E = h ν.
λ ν = v = c (for a photon).
∴ E = h c / λ.

E   = (4.13567 × 10⁻¹⁵ eV s) × (29979245800 cm / s) / (5000 Å) × (1 Å/10⁻⁸ cm).
    = 2.479686 × 10⁷ × 10⁻¹⁵ × 10⁸ eV s cm Å /s / Å / cm
    = 2.480 eV.

1 Rydberg = 13.606 eV.

E   = 2.480 eV × (1 Rydberg/13.606 eV)
    = 0.1823 rydberg.