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Fusion and Stars · Paper 2 — Free Physics SL Practice Questions

1FoundationSAQ-SNuclear fusion and energy production in stars7 marksPaper 2~11 min

Consider the fusion of deuterium (

^2_1\text{H}

) and tritium (

^3_1\text{H}

) in a reactor:

^2_1\text{H} + ^3_1\text{H} \rightarrow ^4_2\text{He} + ^1_0\text{n}

The binding energies per nucleus are: deuterium = 2.22 MeV, tritium = 8.48 MeV, helium-4 = 28.3 MeV, neutron = 0 MeV.

(a)

Explain why energy is released in this fusion reaction using the concept of binding energy. [2 marks]

(b)

Calculate the energy released per fusion event, in MeV. [2 marks]

(c)

A reactor undergoes

1.5 \times 10^{20}

fusion events per second. Determine the power output of the reactor in watts. [3 marks]

Solutions

2MasterySAQ-SLife cycle of stars6 marksPaper 2~9 min

A Hertzsprung–Russell (H-R) for a star cluster shows a clear main sequence, a red giant branch, and a horizontal branch. One star, designated Star X, has left the main sequence and is now a red giant.

(a)

State the event that causes a star to leave the main sequence. [1 mark]

(b)

Explain the changes in the core of Star X that cause it to become a red giant. [2 marks]

(c)

The core of Star X contracts while the outer layers expand. The core has mass

M = 2.0 \times 10^{29}\,\text{kg}

and its radius decreases from

R_i = 1.0 \times 10^{8}\,\text{m}

R_f = 2.0 \times 10^{7}\,\text{m}

. The gravitational potential energy of a uniform sphere is:

U = -\frac{3GM^2}{5R}

where

G = 6.67 \times 10^{-11}\,\text{N\,m}^2\,\text{kg}^{-2}

. Calculate the change in gravitational potential energy of the core. [3 marks]

Solutions

3ChallengeSAQ-LNuclear fusion and energy production in stars7 marksPaper 2~11 min

Data

k = 1.38 \times 10^{-23}\,\text{J\,K}^{-1}

1\,\text{eV} = 1.60 \times 10^{-19}\,\text{J}

- Mass of proton

= 1.007276\,\text{u}

- Mass of helium-4 nucleus

= 4.001506\,\text{u}

- Mass of positron

= 0.000548\,\text{u}

1\,\text{u} = 931.5\,\text{MeV}\,c^{-2}

- Coulomb barrier for two protons

\approx 0.5\,\text{MeV}

- CNO reaction rate

\propto T^{20}

A star with mass

2.5\,M_\odot

burns hydrogen its core via the CNO cycle, in which carbon-12 acts a catalyst:

^{12}_{6}\text{C} + ^{1}_{1}\text{H} \rightarrow ^{13}_{7}\text{N} + \gamma

^{13}_{7}\text{N} \rightarrow ^{13}_{6}\text{C} + e^{+} + \nu_e

^{13}_{6}\text{C} + ^{1}_{1}\text{H} \rightarrow ^{14}_{7}\text{N} + \gamma

^{14}_{7}\text{N} + ^{1}_{1}\text{H} \rightarrow ^{15}_{8}\text{O} + \gamma

^{15}_{8}\text{O} \rightarrow ^{15}_{7}\text{N} + e^{+} + \nu_e

^{15}_{7}\text{N} + ^{1}_{1}\text{H} \rightarrow ^{12}_{6}\text{C} + ^{4}_{2}\text{He}

The net reaction is:

4\,^{1}_{1}\text{H} \rightarrow ^{4}_{2}\text{He} + 2e^{+} + 2\nu_e + 3\gamma

The star's core temperature is

2.0 \times 10^{7}\,\text{K}

. The average kinetic energy of a proton at this temperature is

E_k = \tfrac{3}{2}kT

(a)

Calculate the average kinetic energy of a proton in the star's core in eV. [2 marks]

(b)

Calculate the energy released, in MeV, in the net CNO cycle reaction. [3 marks]

(c)

Explain why fusion can occur even though the average proton kinetic energy found in (a) is far below the Coulomb barrier. [1 mark]

(d)

The core temperature increases to

3.0 \times 10^{7}\,\text{K}

. Determine the factor by which the CNO energy production rate changes. [1 mark]

Solutions

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