| Date | May 2015 | Marks available | 4 | Reference code | 15M.3.SL.TZ1.15 |
| Level | Standard level | Paper | Paper 3 | Time zone | Time zone 1 |
| Command term | Show that and Suggest | Question number | 15 | Adapted from | N/A |
Question
This question is about a particular star called Barnard’s star.
The peak wavelength in the spectrum of Barnard’s star is 940 nm. The following data are available.
\[\frac{{{\text{apparent brightness of Barnard's star}}}}{{{\text{apparent brightness of the Sun}}}} = 2.5 \times {10^{ - 14}}\]
\[\frac{{{\text{luminosity of Barnard's star}}}}{{{\text{luminosity of the Sun}}}} = 3.8 \times {10^{ - 3}}\]
(i) Show that the surface temperature of Barnard’s star is about 3000 K.
(ii) Suggest why Barnard’s star is not likely to be either a white dwarf or a red giant.
(i) Determine, in astronomical units (AU), the distance between Earth and Barnard’s star.
(ii) Calculate the parallax angle for Barnard’s star as observed from Earth.
(iii) Outline how the parallax angle is measured.
Markscheme
(i) \(T = \frac{{0.0029}}{\lambda }\);
3080/3090 (K); (more than 1 SD must be shown)
(ii) temperature too low for white dwarf;
not luminous enough for red giant;
(i) \(L = 4\pi {d^2}b\);
\(\frac{{{d_B}}}{{{d_S}}}\left( { = \sqrt {\frac{{{L_B}}}{{{L_S}}}\frac{{{b_S}}}{{{b_B}}}} } \right) = \sqrt {\frac{{3.8 \times {{10}^{ - 3}}}}{{2.5 \times {{10}^{ - 14}}}}} \);
3.9×105 AU;
(ii) conversion of AU to 1.89 pc;
0.53 (arc-seconds);
with respect to fixed background;
with six months between readings;
parallax angle is half the total angle / OWTTE;
May be shown in a diagram.
