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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)
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