| Date | May 2016 | Marks available | 3 | Reference code | 16M.2.HL.TZ0.8 |
| Level | Higher level | Paper | Paper 2 | Time zone | Time zone 0 |
| Command term | Determine and State | Question number | 8 | Adapted from | N/A |
Question
In beta minus (β−) decay a d quark decays into a u quark, an electron and an electron antineutrino.
Show that lepton number is conserved in this decay.
A nucleus of phosphorus-32 \(\left( {{}_{15}^{32}{\rm{P}}} \right)\) decays by beta minus (β−) decay into a nucleus of sulfur-32 \(\left( {{}_{16}^{32}{\rm{S}}} \right)\). The binding energy per nucleon of \({}_{15}^{32}{\rm{P }}\) is 8.398 MeV and for \({}_{16}^{32}{\rm{S }}\) it is 8.450 MeV.
(i) State what is meant by the binding energy of a nucleus.
(ii) Determine the energy released in this decay.
Quarks were hypothesized long before their existence was experimentally verified. Discuss the reasons why physicists developed a theory that involved quarks.
Markscheme
«lepton number on» LHS = 0 and «lepton number on» RHS = 0 +1−1
OR
quarks have no/0/zero lepton number and the lepton number for electron and the antineutrino cancel
(i)
energy released when a nucleus forms from constituent nucleons
OR
minimum energy needed/work done to break a nucleus up into its constituent nucleons
Do not allow reference to “atom”.
Award [0] for “energy to assemble nucleus”.
Do not allow “particles”, “constituents” or “components” for “nucleons”.
(ii)
«energy/mass difference =» 8.450 – 8.398 «= 0.052 MeV»
Q = 1.7 or 1.66 or 1.664 MeV
OR
2.66 × 10–13 J
quark theory is simpler OR Occam’s razor example OR simple model explains complex observations
quotes experiment that led to quark theory, eg deep inelastic scattering or electron scattering
model incorporates strong/weak interactions/forces between protons and neutrons
model incorporates conservation rules
model explains differences between neutrons and protons OR explains decay of neutron to proton
