(i) Explain why fusion, combining two smaller nuclei into a larger nucleus, releases vast amounts of energy. Use section 36 of the data booklet.

(ii) Outline one advantage of fusion as a source of energy.

Radioactive phosphorus, ³³P, has a half-life of 25.3 days.

(i) Calculate ³³P decay constant λ and state its unit. Use section 1 of the data booklet.

(ii) Determine the fraction of the ³³P sample remaining after 101.2 days.

i

product has higher binding energy «per nucleon»/more stable
OR
nucleons in product more tightly bound «with one another»

lighter elements «than Fe» can fuse/combine with loss of mass/mass defect «and release vast amount of energy»

Accept “mass is converted to energy” for M2

ii

Any one of:
deuterium/fuel is abundant/cheap
«helium» products not radioactive
fusion much less dangerous than fission
large amounts/shipments of radioactive fuel not required
far less radioactive waste «created by fast moving neutrons» has to be stored 

Accept “reduces greenhouse gas emissions/global warming” OR “no radioactive waste” OR “more reliable power” OR “fewer safety issues”.
Do not accept “gives out a large amount of energy” as it is in the stem of the question.

i
«\(\lambda  = \frac{{\ln 2}}{{{t_{\frac{1}{2}}}}} = \frac{{0.693}}{{25.3\,{\text{days}}}} = \)» 2.74 × 10⁻² day⁻¹
Need correct unit for mark.

ii
«4 half-lives; 1 →\(\frac{1}{2}\)→\(\frac{1}{4}\)→\(\frac{1}{8}\)→\(\frac{1}{16}\) =» \(\frac{1}{16}\) / 6.25 × 10⁻²
OR
«\(\frac{N}{{{N_0}}} = {e^{ - {\lambda _t}}} = {e^{ - 0.0274\,\, \times \,\,101.2}} = \)» 6.25 × 10⁻²

Accept 6.25%.