Date | May 2019 | Marks available | 1 | Reference code | 19M.3.sl.TZ1.12 |
Level | SL | Paper | 3 | Time zone | TZ1 |
Command term | Calculate | Question number | 12 | Adapted from | N/A |
Question
Uranium-235, 235U, is bombarded with a neutron causing a fission reaction.
Two products of the fission of 235U are 144Ba and 89Kr.
Write the nuclear equation for this fission reaction.
Outline why the reaction releases energy.
The critical mass for weapons-grade uranium can be as small as 15 kg. Outline what is meant by critical mass by referring to the equation in (a)(i).
The daughter product, 89Kr, has a half-life of 3.15 min.
Calculate the time required, in minutes, for the mass of 89Kr to fall to 6.25 % of its initial value.
Markscheme
235U + 1n → 144Ba + 89Kr + 3 1n [✔]
greater binding energy per nucleon in products than reactant [✔]
Note: Accept “mass of products less than reactants” OR “mass converted to energy/E = mc2”.
mass/amount/quantity required so that «on average» each fission/reaction results in a further fission/reaction [✔]
at least one of the «3» neutrons produced must cause another reaction [✔]
Note: Accept “minimum mass of fuel needed for the reaction to be self-sustaining”.
«6.25 % = 4 half-lives, so 4 × 3.15 =» 12.6 «min» [✔]
Examiners report
Only about a third of the candidates could write the nuclear equation for the requested fission reaction.
Only about a quarter of the candidates could explain the release of energy, usually in terms of the change in nuclear binding energy.
Less than half the candidates could correctly explain “critical mass” with even fewer combining the definition as the minimum mass for a sustainable fission reaction with the required clarification in terms of neutrons creating a chain reaction.
Another question that created far more problems than anticipated with only about half gaining the mark. Many students appeared not to realise that 6.25 % is 1/16, hence the amount remaining after 4 half lives.