| Date | May 2017 | Marks available | 3 | Reference code | 17M.3.HL.TZ1.5 |
| Level | Higher level | Paper | Paper 3 | Time zone | Time zone 1 |
| Command term | Determine | Question number | 5 | Adapted from | N/A |
Question
A proton is accelerated from rest through a potential difference V to a speed of 0.86c.
Calculate the potential difference V.
[3]
a.
The proton collides with an antiproton moving with the same speed in the opposite direction. As a result both particles are annihilated and two photons of equal energy are produced.
Determine the momentum of one of the photons.
[3]
b.
Markscheme
γ = 1.96
Ek = (γ − 1) m0c2 = 900 «Me\(\,\)V»
pd ≈ 900 «MV»
Award [2 max] if Energy and Potential difference are not clearly distinguished, eg by the unit.
[3 marks]
a.
energy of proton = γmc2 = 1838 «Me\(\,\)V»
total energy available = energy of proton + energy of antiproton = 1838 + 1838 = 3676 «Me\(\,\)V»
momentum of a one photon = Total energy / 2c = 1838 «Me\(\,\)Vc–1»
[3 marks]
b.
Examiners report
[N/A]
a.
[N/A]
b.
Syllabus sections
Option A: Relativity » Option A: Relativity (Additional higher level option topics) » A.4 – Relativistic mechanics (HL only)
Show 41 related questions
