DP Physics (first assessment 2025)
Question 23M.2.HL.TZ1.5bi
| Date | May 2023 | Marks available | [Maximum mark: 2] | Reference code | 23M.2.HL.TZ1.5bi |
| Level | HL | Paper | 2 | Time zone | TZ1 |
| Command term | Show that | Question number | i | Adapted from | N/A |
i.
[Maximum mark: 2]
23M.2.HL.TZ1.5bi
(i)
Show that the energy released is about 18 MeV.
[2]
Markscheme
«𝜇» = 2.0141 + 3.0160 − (4.0026 + 1.008665) «= 0.0188 u»
OR
In MeV: 1876.13415 + 2809.404 − (3728.4219 + 939.5714475) ✓
= 0.0188 × 931.5 OR = 17.512 «MeV» ✓
Must see either clear substitutions or answer to at least 3 s.f. for MP2.
