Date | May 2011 | Marks available | 4 | Reference code | 11M.3.SL.TZ1.14 |
Level | Standard level | Paper | Paper 3 | Time zone | Time zone 1 |
Command term | Show that | Question number | 14 | Adapted from | N/A |
Question
This question is about the Hertzsprung–Russell (HR) diagram and using it to determine some properties of stars.
The diagram below shows the grid of a HR diagram, on which the positions of selected stars are shown. (LS = luminosity of the Sun.)
(i) Draw a circle around the stars that are red giants. Label this circle R.
(ii) Draw a circle around the stars that are white dwarfs. Label this circle W.
(iii) Draw a line through the stars that are main sequence stars.
Explain, without doing any calculation, how astronomers can deduce that star B has a larger diameter than star A.
Using the following data and information from the HR diagram, show that star A is at a distance of about 800 pc from Earth.
Apparent brightness of the Sun =1.4×103Wm−2
Apparent brightness of star A = 4.9×10−9Wm−2
Mean distance of Sun from Earth =1.0 AU
1 pc = 2.1×105AU
Explain why the distance of star A from Earth cannot be determined by the method of stellar parallax.
Markscheme
(i) circle labelled R as shown above;
Accept answers that include the star B within the circle.
(ii) circle labelled W as shown above;
(iii) any line (not necessarily straight) going from top left to bottom right, through or near all or most of stars;
star B has lower temperature;
star B has (slightly) larger luminosity / stars have approximately same luminosity;
surface area calculated from L=σAT4, so star B has larger surface area/diameter / to give the same/similar luminosity at lower temperature, star B must have bigger diameter/surface area;
(from HR diagram) LA =105LS;
\(b = \frac{L}{{4\pi {d^2}}}\) used;
to give \(\frac{{{d_{\rm{A}}}}}{{{d_{\rm{S}}}}} = \sqrt {\frac{{{L_{\rm{A}}}}}{{{L_{\rm{S}}}}} \times \frac{{{b_{\rm{S}}}}}{{{b_{\rm{A}}}}}} = \sqrt {{{10}^5} \times \frac{{1.4 \times {{10}^3}}}{{4.9 \times {{10}^{ - 9}}}}} \);
hence dA =1.7×108 AU;
= 800 pc
Do not award a mark for the conversion from AU to pc.
the parallax angle is too small to be measured accurately / the distance is greater than the limit for stellar parallax, which is 100 pc;
Accept any value from 100–800 pc for limit. Do not accept “it’s too far away”.