Question 22N.2.HL.TZ0.6
| Date | November 2022 | Marks available | [Maximum mark: 13] | Reference code | 22N.2.HL.TZ0.6 |
| Level | HL | Paper | 2 | Time zone | TZ0 |
| Command term | Calculate, Discuss, Outline, Show | Question number | 6 | Adapted from | N/A |
Outline, by reference to nuclear binding energy, why the mass of a nucleus is less than the sum of the masses of its constituent nucleons.
[2]
according to ΔE = Δmc2 / identifies mass energy equivalence ✓
energy is released when nucleons come together / a nucleus is formed «so nucleus has less mass than individual nucleons»
OR
energy is required to «completely» separate the nucleons / break apart a nucleus «so individual nucleons have more mass than nucleus» ✓
Accept protons and neutrons.
Still several answers that thought that the nucleus needed to gain energy to bind it together. Most candidates scored at least one for recognising some form of mass/energy equivalence, although few candidates managed to consistently express their ideas here.
Polonium-210 (Po-210) decays by alpha emission into lead-206 (Pb-206).
The following data are available.
Nuclear mass of Po-210 = 209.93676 u
Nuclear mass of Pb-206 = 205.92945 u
Mass of the alpha particle = 4.00151 u
Calculate, in MeV, the energy released in this decay.
[2]
(mpolonium − mlead − mα)c2 OR (209.93676 − 205.92945 − 4.00151)
OR
mass difference = 5.8 × 10−3 ✓
conversion to MeV using 931.5 to give 5.4 «MeV» ✓
Allow ECF from MP1.
Award [2] for a BCA.
Award [1] for 8.6 x 10−13 J.
Generally, well answered. There were quite a few who fell into the trap of multiplying by an unnecessary c2 as they were not sure of the significance of the unit of u.
The polonium nucleus was stationary before the decay.
Show, by reference to the momentum of the particles, that the kinetic energy of the alpha particle is much greater than the kinetic energy of the lead nucleus.
[3]
ALTERNATIVE 1
energy ratio expressed in terms of momentum, e.g. ✓
hence ✓
«so has a much greater KE»
OR
«much» greater than «so has a much greater KE» ✓
ALTERNATIVE 2
alpha particle and lead particle have equal and opposite momenta ✓
so their velocities are inversely proportional to mass ✓
but KE ∝ v2 «so has a much greater KE» ✓
Those who answered using the mass often did not get MP3 whereas those who converted to the number of particles or moles before the first calculation did, although that could be considered an unnecessary complication.
In the decay of polonium−210, alpha emissions can be accompanied by the emissions of gamma photons, all of the same wavelength of 1.54 × 10−12 m.
Discuss how this observation provides evidence for discrete nuclear energy levels.
[3]
photon energy is determined by its wavelength ✓
photons are emitted when nucleus undergoes transitions between its «nuclear» energy levels
OR
photon energy equals the difference between «nuclear» energy levels ✓
photons have the same energy / a fixed value
OR
energy is quantized / discrete ✓
Many identified conservation of momentum and consequently the relative velocities but it was common to miss MP3 for correctly relating this to KE.
A sample contains 5.0 g of pure polonium-210. The decay constant of polonium-210 is 5.8 × 10−8 s−1. Lead-206 is stable.
Calculate the mass of lead-206 present in the sample after one year.
[3]
undecayed mass « g» ✓
mass of decayed polonium « undecayed mass» «g» ✓
mass of lead «» «g» ✓
Allow [2] max for answers that ignore mass difference between Pb and Po (4.2 g).
Allow calculations in number of particles or moles for MP1 and MP2.
Allow ECF from MP1 and MP2.
Several answers referred incorrectly to electron energy levels. Successful candidates managed to score full marks, although it was also common to miss the relationship between energy and wavelength.