IB Physics HL E.3.4 Mass-energy equivalence Question Bank

Practise IB Physics SL/HL E.3.4 by applying mass-energy equivalence concepts to exam-style physics questions.

Exam points

identify the relevant physical quantity, model or diagram
apply the correct equation and units to the evidence
justify the result using conservation laws, fields or energy relationships

[Maximum number: 2]

The diagram shows a simplified energy-balance model for the Earth surface-atmosphere system.

Figure for Question E.3.4—Mass-energy equivalence question 1 — IB Physics HL

The following data are given:

\begin{aligned} \text { Average albedo of Earth } & =0.30 \\ \text { Average global temperature of the surface } & =288 \mathrm{~K} \\ \text { Average Earth-Sun distance } & =1.5 \times 10^{11} \mathrm{~m} \end{aligned}

The primary energy source of the Sun is the proton-proton (p-p) chain of fusion reactions. Four protons and two electrons produce a helium nucleus together with neutrinos and gamma photons. The overall reaction is:

4{ }_{1}^{1} \mathrm{p}+2{ }_{-1}^{0} \mathrm{e} \rightarrow{ }_{2}^{4} \mathrm{He}+2{ }_{0}^{0} v_{e}+4 \gamma

The mass of the helium nucleus is 4.001506 u . Calculate, in MeV , the energy released in the reaction.

ALTERNATIVE 1
$\begin{aligned} \Delta m=4 \times 1.007276+2 \times 0.000549-4.001506 «=0.028696 \mathrm{u} » \text { «0.028696×931.5 }=» 26.7 \mathrm{MeV} \text { ALTERNATIVE } 2 \begin{array}{l} \Delta m=4 \times 938+2 \times 0.511-4.001506 \times 931.5 \ll=25.6 \mathrm{MeV} \mathrm{c}^{-2} \gg 25.6 \mathrm{MeV} \end{array} \end{aligned}$

Marks can only be awarded from one Alternative

Unit is not required.

Particle masses in u and $M e V c^{-2}$ lead to differently rounded answers, working must be checked.

Award [1] max if electrons are omitted. This leads to ALT $125.7 \ll \mathrm{MeV} \gg$ , ALT $224.6 \ll \mathrm{MeV} \gg$

Award ECF if incorrect number of particles used e.g. 1 electron rather than 2.

All question bank results loaded

E.3.4—Mass-energy equivalence question 1