DP Physics (last assessment 2024)
Question 18M.2.SL.TZ2.6b.ii
| Date | May 2018 | Marks available | [Maximum mark: 1] | Reference code | 18M.2.SL.TZ2.6b.ii |
| Level | SL | Paper | 2 | Time zone | TZ2 |
| Command term | Show that | Question number | b.ii | Adapted from | N/A |
b.ii.
[Maximum mark: 1]
18M.2.SL.TZ2.6b.ii
Rhodium-106 () decays into palladium-106 () by beta minus (β–) decay.
The binding energy per nucleon of rhodium is 8.521 MeV and that of palladium is 8.550 MeV.
(b.ii)
Show that the energy released in the β– decay of rhodium is about 3 MeV.
[1]
Markscheme
Q = 106 × 8.550 − 106 × 8.521 = 3.07 «MeV»
«Q ≈ 3 Me V»
[1 mark]
