Sketch, on the axes in Figure 1, the black-body radiation curve for a typical star - AQA - A-Level Physics - Question 2 - 2018 - Paper 4

To determine the black-body temperature from the curve, we can use Wien's displacement law, which states that the peak wavelength ( ( \lambda_{peak} ) ) is inversely proportional to the temperature (T) of the black body:

$\lambda_{peak} = \frac{b}{T}$

where b is Wien's displacement constant (approximately 2898 μm·K).

By measuring the peak wavelength from the sketch, we can rearrange the formula to find the temperature:

$T = \frac{b}{\lambda_{peak}}$

where ( T ) is in Kelvin.

To evaluate the suggestion that 61 Cygnus A and B form a binary system, we must compare their apparent magnitudes and brightness as perceived from Earth.

1. Comparison of Brightness:

   • The apparent magnitude of 61 Cygnus A is 5.2, while for 61 Cygnus B it is 6.1. A lower apparent magnitude indicates a brighter star. Hence, Cygnus A appears brighter than Cygnus B.

2. Distance:

   • Assuming the distance to both stars is roughly the same, we can calculate their absolute magnitudes and luminosities based on their apparent magnitudes. The formulas are:

   $M = m - 5 \cdot (\log_{10}(d) - 1)$

   for absolute magnitude (where d is the distance in parsecs). Using these values, we can evaluate their luminosities, using:

   $L \propto 10^{-0.4M}$

3. Conclusion:

   • If the brightness calculations yield similar values, it supports the idea that they are part of a binary system. If Cygnus A appears significantly brighter, it suggests they may not be at the same distance, challenging the binary system argument.

The spectral class of 61 Cygnus A corresponds to a temperature of 4500 K. Based on the classifications, a temperature around this range typically categorizes a star in the 'K' spectral class, which indicates it is a cooler star. Thus, when asked to tick the correct box, 'K' should be selected.