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Table 2 summarises some of the properties of four stars in the constellation Hercules - AQA - A-Level Physics - Question 3 - 2017 - Paper 4

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Table 2 summarises some of the properties of four stars in the constellation Hercules. Table 2 | Star | Distance/pc | Spectral class | Apparent magnitude |... show full transcript

Worked Solution & Example Answer:Table 2 summarises some of the properties of four stars in the constellation Hercules - AQA - A-Level Physics - Question 3 - 2017 - Paper 4

Step 1

Define the parsec. You may use a diagram as part of your answer.

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Answer

A parsec is a unit of distance used in astronomy, equivalent to the distance at which one astronomical unit (AU) subtends an angle of one arcsecond ("1 arcsec"). In simpler terms, 1 parsec is approximately 3.26 light-years. A diagram can illustrate this concept, demonstrating that at a distance of 1 parsec, the angle from the position of the Earth to a star is 1 arcsecond when observed from a distance of 1 AU.

Step 2

Deduce which star is larger, Korophoros or Rutilicus.

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Answer

To deduce which star is larger between Korophoros and Rutilicus, we need to examine their properties. The star that is of spectral class O tends to be the largest, while the spectral class G (Korophoros) indicates a medium-sized star. Rutilicus is classified as a B-type star, which can also be larger than G-type stars. Given Rutilicus's spectral classification, it is generally inferred to be larger than Korophoros.

Step 3

One of the four stars has the peak in its black-body radiation curve at a wavelength of 1.0 μm. Calculate the corresponding temperature for this curve.

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To find the corresponding temperature for a black-body radiation curve peaking at a wavelength of 1.0 μm, we use Wien's Displacement Law, which states: extT=bλmax ext{T} = \frac{b}{\lambda_{max}} where (b) is a constant approximately equal to 2898 μm·K. Substituting in this equation: extT=28981.0=2898K ext{T} = \frac{2898}{1.0} = 2898 \,\text{K}.

Step 4

Explain which star produced the black-body radiation curve described in question 03.3.

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The star that produces the black-body radiation curve peaking at 1.0 μm is likely Rasalgethi, as it is a M-type star. M-type stars have cooler temperatures compared to other spectral classes and have their peak emissions in the infrared spectrum, tending towards longer wavelengths. Thus, a peak at 1.0 μm aligns with the characteristics of an M-type star.

Step 5

Which star has the brightest absolute magnitude? Tick (✓) the correct box.

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Answer

To determine which star has the brightest absolute magnitude, we examine the apparent magnitudes in conjunction with their distances. Rutilicus has the lowest apparent magnitude at 2.1, suggesting it could be the brightest when factoring in distance. Therefore:

  • Rutilicus ✓

Step 6

Determine the absolute magnitude of Sarin.

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Answer

The absolute magnitude can be calculated using the formula: M=m5(log10(d)1)M = m - 5 \cdot (\log_{10}(d) - 1) Where (M) is the absolute magnitude, (m) is the apparent magnitude, and (d) is the distance in parsecs. For Sarin:

  • Apparent magnitude (m) = 3.1
  • Distance (d) = 23 pc

Substituting into the formula: M=3.15(log10(23)1)M = 3.1 - 5 \cdot (\log_{10}(23) - 1) Calculating (\log_{10}(23) \approx 1.362), we find: M=3.15(1.3621)3.150.3623.11.811.29M = 3.1 - 5 \cdot (1.362 - 1) \approx 3.1 - 5 \cdot 0.362 \approx 3.1 - 1.81 \approx 1.29 Thus, the absolute magnitude of Sarin is approximately 1.29.

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