3C 273 was the first quasar to be discovered - AQA - A-Level Physics - Question 3 - 2022 - Paper 4

To determine the absolute magnitude, we use the formula: $M = m - 5 imes ext{log}_{10}(d) + 5$ Where:

• M is the absolute magnitude,
• m is the apparent magnitude (12.8), and
• d is the distance in parsecs (760 Mpc = 760,000,000 parsecs).

Plugging in the numbers: $M = 12.8 - 5 imes ext{log}_{10}(760000000) + 5$ Calculating ext{log}_{10}(760000000), we find: $M = 12.8 - 5 imes 8.880 + 5$ $M ext{ resulting in approximately } -27.$

Quasar 3C 273 would be the brighter object due to its more negative absolute magnitude compared to IC 1101. In celestial terms, a lower (more negative) absolute magnitude indicates a higher intrinsic brightness.

Using the absolute magnitudes:

• For quasar: M = -27
• For IC 1101: M = -22.8

The difference in absolute magnitudes is: $ext{Difference} = -22.8 - (-27) = 4.2$

Using the formula for brightness ratio: rac{B_1}{B_2} = 10^{0.4 imes ext{Difference}}

Calculating: rac{B_1}{B_2} = 10^{0.4 imes 4.2} ext{ which is approximately } 2.51^{3.33}.

The volume of the black hole's event horizon can be calculated using the formula for the volume of a sphere: $V = \frac{4}{3} \pi r^3$

Where the radius r can be derived from the mass of the black hole: $r = \frac{2GM}{c^2}$ Substituting in values:

• G (gravitational constant) is approximately $6.67 imes 10^{-11} ext{ m}^3/ ext{kg s}^2$,
• M = $7.1 imes 10^{11} M_⊙$ (where $M_⊙$ mass of the Sun is $1.989 imes 10^{30}$ kg).

Calculating: $M = 7.1 imes 10^{11} imes (1.989 imes 10^{30}) ext{ kg}$

Then substituting values to find the volume and density: $\text{average density} = \frac{\text{mass}}{V}$

This will yield the average density in kg/m³.