Question 19M.3.sl.TZ1.12
| Date | May 2019 | Marks available | [Maximum mark: 5] | Reference code | 19M.3.sl.TZ1.12 |
| Level | sl | Paper | 3 | Time zone | TZ1 |
| Command term | Calculate, Outline, Write | Question number | 12 | Adapted from | N/A |
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.
[1]
235U + 1n → 144Ba + 89Kr + 3 1n [✔]
Only about a third of the candidates could write the nuclear equation for the requested fission reaction.
Outline why the reaction releases energy.
[1]
greater binding energy per nucleon in products than reactant [✔]
Note: Accept “mass of products less than reactants” OR “mass converted to energy/E = mc2”.
Only about a quarter of the candidates could explain the release of energy, usually in terms of the change in nuclear binding 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).
[2]
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”.
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.
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.
[1]
«6.25 % = 4 half-lives, so 4 × 3.15 =» 12.6 «min» [✔]
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.