Question 23M.2.HL.TZ2.9
| Date | May 2023 | Marks available | [Maximum mark: 11] | Reference code | 23M.2.HL.TZ2.9 |
| Level | HL | Paper | 2 | Time zone | TZ2 |
| Command term | Calculate, Determine, Estimate, Explain, Show that, State | Question number | 9 | Adapted from | N/A |
Magnesium-27 nuclei () decay by beta-minus (β−) decay to form nuclei of aluminium-27 (Al).
Show, using the data, that the energy released in the decay of one magnesium-27 nucleus is about 2.62 MeV.
Mass of aluminium-27 atom = 26.98153 u
Mass of magnesium-27 atom = 26.98434 u
The unified atomic mass unit is 931.5 MeV c−2.
[1]
(26.98434 - 26.98153) × 931.5
OR
2.6175 «MeV» seen ✓
A Magnesium-27 nucleus can decay by one of two routes:
Route 1: 70% of the beta particles are emitted with a maximum kinetic energy of 1.76656 MeV, accompanied by a gamma photon of energy 0.84376 MeV.
Route 2: 30% of the beta particles have a maximum kinetic energy of 1.59587 MeV with a gamma photon of energy 1.01445 MeV.
The final state of the aluminium-27 nucleus is the same for both routes.
State the conclusion that can be drawn from the existence of these two routes.
[1]
evidence for nuclear energy levels ✓
Calculate the difference between the magnitudes of the total energy transfers in parts (a) and (b).
[1]
Difference = 2.6175 – (1.76656 +0.84376) = 2.6175 – 2.61032 = 0.007195 «MeV»
OR
Difference = 2.6175 – (1.59587 +1.01445) = 2.61032 = 0.007195 «MeV» ✓
Explain how the difference in part (b)(ii) arises.
[1]
Another particle/«anti» neutrino is emitted «that accounts for this mass / energy» ✓
Small amounts of magnesium in a material can be detected by firing neutrons at magnesium-26 nuclei. This process is known as irradiation.
Magnesium-27 is formed because of irradiation. The products of the beta-particle emission are observed as the magnesium-27 decays to aluminium-27.
The smallest mass of magnesium that can be detected with this technique is 1.1 × 10−8 kg.
Show that the smallest number of magnesium atoms that can be detected with this technique is about 1017.
[2]
So 1.1 × 10−8 kg ≡ × 10−8 «mol»
OR
Mass of atom = 27 × 1.66 × 10−27 «kg» ✓
2.4 − 2.5 x 1017 atoms ✓
A sample of glass is irradiated with neutrons so that all the magnesium atoms become magnesium-27. The sample contains 9.50 × 1015 magnesium atoms.
The decay constant of magnesium-27 is 1.22 × 10−3 s−1.
Determine the number of aluminium atoms that form in 10.0 minutes after the irradiation ends.
[3]
N10 = 9.50 × 1015 × e−0.00122×60 seen ✓
N10 = 4.57 × 1015 ✓
So number of aluminium-27 nuclei = (9.50 – 4.57) × 1015 = 4.9(3) × 1015 ✓
Estimate, in W, the average rate at which energy is transferred by the decay of magnesium-27 during the 10.0 minutes after the irradiation ends.
[2]
Total energy released = ans (c)(ii) × 2.62 × 106 × 1.6 × 10−19 «= 2100 J» ✓
« =» 3.4 −3.5 «W» ✓