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Date May 2011 Marks available 3 Reference code 11M.3.SL.TZ1.14
Level Standard level Paper Paper 3 Time zone Time zone 1
Command term Deduce 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.

 

[3]
a.

Explain, without doing any calculation, how astronomers can deduce that star B has a larger diameter than star A.

 

[3]
b.

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

 

[4]
c.

Explain why the distance of star A from Earth cannot be determined by the method of stellar parallax.

[1]
d.

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;

a.

star B has lower temperature;
star B has (slightly) larger luminosity / stars have approximately same luminosity;
surface area calculated from LAT4, 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;

b.

(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.

c.

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”.

d.

Examiners report

[N/A]
a.
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b.
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c.
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d.

Syllabus sections

Option D: Astrophysics » Option D: Astrophysics (Core topics) » D.1 – Stellar quantities
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