3C 273 was the first quasar to be discovered - AQA - A-Level Physics - Question 3 - 2022 - Paper 4
Question 3
3C 273 was the first quasar to be discovered.
IC 1101 is one of the largest galaxies known.
Table 2 shows some information about these objects.
Table 2
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Worked Solution & Example Answer:3C 273 was the first quasar to be discovered - AQA - A-Level Physics - Question 3 - 2022 - Paper 4
Step 1
State the property of the quasar that led to its discovery.
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Answer
The quasar 3C 273 is known for being a high power and powerful radio emitter, which was a significant property leading to its discovery.
Step 2
Show that the absolute magnitude X of quasar 3C 273 is about -27.
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Answer
To determine the absolute magnitude X, we use the formula that relates absolute and apparent magnitudes to distance:
M=m−5imes(extlog10(D)−1)
Where:
M is the absolute magnitude
m is the apparent magnitude (12.8 for 3C 273)
D is the distance in parsecs (760 Mpc = 760 × 10^6 pc)
Now substituting the values:
M=12.8−5imes(extlog10(760imes106)−1)
Calculating gives:
Thus,
$$M ext{ approximately equals } 12.8 - 5 imes (8.880 - 1) ext{ gives us } M ext{ around } -27.$$
Step 3
Explain which would be the brighter object.
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Answer
The quasar 3C 273 will be brighter than galaxy IC 1101 because it has a more negative absolute magnitude. In astronomy, a lower (or more negative) absolute magnitude indicates a brighter object. Given the absolute magnitudes of -27 for 3C 273 and -22.8 for IC 1101, quasar 3C 273 is indeed the brighter object.
Step 4
Go on to calculate the ratio brightness of brighter object brightness of dimmer object.
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The ratio of brightness can be calculated using the formula:
extRatio=10(m2−m1)/5
Where:
m1 is the apparent magnitude of the brighter object (3C 273), which is 12.8.
m2 is the apparent magnitude of the dimmer object (IC 1101), which is 14.7.
Substituting the values:
extRatio=10(14.7−12.8)/5=101.9/5extorapproximately2.51
Step 5
Calculate the average density within the event horizon of the black hole.
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Answer
To find the average density within the event horizon of the black hole, we need to use the formula:
ho = \frac{M}{V}$$
Where:
- $
ho$ is the average density,
- $M$ is the mass of the black hole ($7.1 imes 10^{11} M_S$),
- $V$ is the volume of the black hole, calculated for a sphere:
$$V = \frac{4}{3} \pi r^3$$
To find the radius $r$ use the Schwarzschild radius:
$$r_s = \frac{2GM}{c^2}$$
Substitute $M = 7.1 imes 10^{11} M_S$ and use $M_S = 1.989 × 10^{30} kg$ for the sun's mass. Calculate volume and then density:
$$ \rho = \frac{7.1 imes 10^{11} imes 1.989 × 10^{30} kg}{ \frac{4}{3} \pi (r_s)^3}$$
Calculating will yield the average density within the event horizon.