(i) The radius of star X is 4.5 R_S where R_S is the radius of the Sun. The surface temperature of the Sun is 5.7×10³K.

Determine the ratio \(\frac{{{\rm{luminosity of star X}}}}{{{\rm{luminosity of the Sun}}}}\).

(ii) Calculate, assuming that the power in the mass–luminosity relationship is 3.5, the ratio \(\frac{{{\rm{mass of star X}}}}{{{\rm{mass of Sun}}}}\).

On the Hertzsprung–Russell diagram, label

(i) the position of star X with the letter X.
(ii) the position of the Sun with the letter S.

Explain, with reference to the Chandrasekhar limit, whether or not star X will become a white dwarf.

(i) \(\frac{{{L_X}}}{{{L_S}}} = \frac{{\sigma {r_X}^2{T_X}^4}}{{\sigma {r_S}^2{T_S}^4}}\);
\( = \frac{{{{4.5}^2} \times {{9700}^4}}}{{{{5700}^4}}}\);
=170;
Accept answers that use T=10000 (K) to give an answer of 190.

(ii) \(\frac{{{M_X}}}{{{M_S}}} = {\left[ {\frac{{{L_X}}}{{{L_S}}}} \right]^{\frac{1}{{3.5}}}}\);
\( = {\left[ {170} \right]^{\frac{1}{{3.5}}}}\);
=4.3;
Award [3] for a bald correct answer.

Chandrasekhar limit below which white dwarf stars can exist is 1.4 solar masses;
X is not going to form a white dwarf / X will not form a white dwarf if its final core/remnant mass is greater than the Chandrasekhar limit; 
Allow ECF from (b)(ii).

There was a lot of poor algebra and messy working. There was a much poorer understanding of ratio in this question, in particular dealing with the inverse power of 3.5.

was well done in general.

was well done in general, but many candidates thought that the Chandrasekhar limit was 4 M_S. The question was slightly confusing because the mass worked out in (b)(ii) is for a main sequence star, not a remnant after a supernova. The markscheme was flexible enough to allow for candidates who took the calculated mass to be the mass of a remnant, or those who said that the star would form a white dwarf if the remnant mass was less than the Chandrasekhar limit.