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

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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 | Object ... show full transcript

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=m5imes(extlog10(D)1)M = m - 5 imes ( ext{log}_{10}(D) - 1) Where:

  • MM is the absolute magnitude
  • mm is the apparent magnitude (12.8 for 3C 273)
  • DD is the distance in parsecs (760 Mpc = 760 × 10^6 pc)

Now substituting the values: M=12.85imes(extlog10(760imes106)1)M = 12.8 - 5 imes ( ext{log}_{10}(760 imes 10^6) - 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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Answer

The ratio of brightness can be calculated using the formula: extRatio=10(m2m1)/5 ext{Ratio} = 10^{(m_2 - m_1) / 5} Where:

  • m1m_1 is the apparent magnitude of the brighter object (3C 273), which is 12.8.
  • m2m_2 is the apparent magnitude of the dimmer object (IC 1101), which is 14.7.

Substituting the values: extRatio=10(14.712.8)/5=101.9/5extorapproximately2.51 ext{Ratio} = 10^{(14.7 - 12.8) / 5} = 10^{1.9 / 5} ext{ or approximately } 2.51

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.

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