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

To find the absolute magnitude $M$ of quasar 3C 273, we use the formula:

$M = m - 5 imes ext{log}_{10}(d/10)$

Given:

• Apparent magnitude $m = 12.8$
• Distance $d = 760$ Mpc

Calculating:

• Convert distance to parsecs: $d = 760 imes 10^6$ pc.
• Substitute into the equation:

$M = 12.8 - 5 imes ext{log}_{10}(760) = 12.8 - 5 imes 2.880$ M utapprle -27

The quasar would be the brighter object because it has a more negative absolute magnitude compared to the galaxy. The brightness is inversely related to the absolute magnitude—more negative values indicate greater intrinsic brightness. In this case, 3C 273 has an absolute magnitude around -27, while IC 1101 has -22.8.

To calculate the ratio of brightness:

$ext{ratio} = 10^{(M_{dimmer} - M_{brighter})/2.5}$

Substituting values: ext{ratio} = 10^{(-22.8 + 27)/2.5} = 10^{4.2/2.5} \\ ext{ratio} utapprle 16.5

The average density $\rho$ can be computed using:

$\rho = \frac{M}{V}$

Where:

• $M = 7.1 \times 10^{11} M_{\odot}$; given $M_{\odot} = 1.989 \times 10^{30} \, \text{kg}$.
• Volume of the event horizon (assuming a Schwarzschild radius) is given by: $V = \frac{4}{3} \\pi R^3$

The Schwarzschild radius $R$ is: $R = 2 \frac{G M}{c^2}$

With constants:

• $G = 6.674 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2}$
• $c = 3 \times 10^8 \, \text{m/s}$

Calculate $R$: $R = 2 \frac{(6.674 \times 10^{-11})(7.1 \times 10^{11} \times 1.989 \times 10^{30})}{(3 \times 10^8)^2} \\ ext{and substitute into } V$ Finally, solve for \rho.