Oxford Revise AQA A Level Physics | Chapter P20 answers

P20: Nuclear energy

Question

Answers

Extra information

Mark

AO

Spec reference

01.1

In path A, the alpha particle is directed at the centre of the nucleus, so the electrostatic force is along the same line that the alpha particle is travelling, but in the opposite direction, so it stops and goes back out

In path B, the force and velocity are not parallel, so the electrostatic force on the alpha particle changes its direction

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3.8.1.5

01.2

\(
\begin{aligned}
& E = \frac{Q_1 Q_2}{4\pi \varepsilon_0 d}\\
& 5 \times 1.6 \times 10^{-13} = \frac{4 \times 79 \times \left( 1.6 \times 10^{-19} \right)^2}{4\pi \times 8.85 \times 10^{-12} \times d}\\
\end{aligned}
\) \(
\begin{aligned}
d & = \frac{4 \times 79 \times \left( 1.6 \times 10^{-19} \right)^2}{4\pi \times 8.85 \times 10^{-12} \times 5 \times 1.6 \times 10^{-13}}\\
& = 9.1 \times 10^{-14}\mathrm{\ m}
\end{aligned}
\)

Use of electric potential energy

Answer

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3.8.1.5

01.3

\(
\begin{aligned}
& \text{Mass of a person = 70 kg}\\
& \text{Density } = \frac{M}{V}\\
& \text{Mass of gold nucleus } = 197 \times 1.667 \times 10^{-27}= 3.28 \times 10^{-25}\mathrm{\ kg}\\
& \text{Volume of nucleus } = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi \left( 9.1 \times 10^{-14} \right)^3 = 3.16 \times 10^{-39}\mathrm{\ m}^3\\
& \text{Nuclear density } = \frac{\text{mass}}{\text{volume}} = \frac{3.28 \times 10^{-25}}{3.16 \times 10^{-39}\mathrm{\ m}^3}\ = 1.04 \times 10^{14}\mathrm{\ kg m}^{-3}\\
& \text{Volume } = \frac{\text{mass}}{\text{density}} = \frac{70}{1.04 \times 10^{14}} = 6.74 \times 10^{-13}\mathrm{\ m}^3\\
& \text{Which is the volume of a cube of sides } 8.8 \times 10^{-5}\mathrm{\ m}
\end{aligned}
\)

human mass estimate between 50 and 100 kg

Calculations of mass and volume of nucleus

Correct volume

Correct cube dimension

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3.8.1.5

01.4

If Z is half, d is halved, so volume is reduced by a factor of (0.5)3

Mass of nucleus is halved, so density of nucleus is reduced by a factor of 4

Volume of person is increased by a factor of 4, and the side of a cube is increased by a factor of 41/3 / 1.5

Effect of Z on d

Effect of A on density

Answer

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3.8.1.5

02.1

The neutron number decreases by 2

The proton number decreases by 2

Both needed for mark

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3.8.1.4

02.2

\(
\begin{aligned}
\text{Change in energy}& = 4.78 – 4.59 = 0.19\mathrm{\ MeV}\\
& = 0.19 \times 1.6 \times 10^{-13}\\
& = 3.04 \times 10^{-14}\mathrm{\ J}\\
\end{aligned}
\) \(
\begin{aligned}
E = \frac{hc}{\lambda},\ \lambda = \frac{hc}{E} & = \frac{6.63 \times {10}^{-34} \times 3.0 \times {10}^8}{3.04 \times 10^{-14}}\\
& = 6.54 \times 10^{-12}\mathrm{\ m}
\end{aligned}
\)

Change in energy

Answer

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3.8.1.4

02.3

In beta decay, a neutron decays to a proton, so the neutron number decreases by 1 and the proton number increases by 1

There are 88 protons in radium-225, and 89 protons in actinium(-226)

There are 137 neutrons in radium-225, and 136 neutrons in actinium(-226)

The arrow goes from (88, 137) to (89, 136)

Correct arrow on diagram:

Explanation of what happens in beta decay

Changes to N and Z

Description of arrow

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3.8.1.4

02.4

Any two from:

  • the arrow starts at (88, 138)
  • the arrow ends at (86, 136)
  • the arrow is twice as long
  • the arrow goes from right to left

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3.8.1.4

03.1

The wavelength should be about the same size as the object producing the pattern

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3.8.1.5

03.2

The ‘target’, in this case the nucleus, is not precisely defined

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3.8.1.5

03.3

Two correct reasons, for example:

Alpha particles contain nucleons that interact in terms of the strong nuclear force

Alpha particles are not point particles

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3.8.1.5

03.4

The measurements of maxima or minima from the graph are not easy to define precisely

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3.8.1.5

04.1

The difference is a measure of the energy needed to separate the nucleons that are bound as a result of the strong nuclear force

Deuterium nucleus contains one proton and one neutron

mproton = 1.00728 u

mneutron = 1.00867 u

mproton + mneutron = 1.007276 + 1.008665 = 2.01595 u

\(
\begin{aligned}
\text{Mass difference} & = 2.397 \times 10^{-3}\mathrm{\ u}\\
& = 2.397 \times 10^{-3}\text{ u }\times 1.661 \times 10^{-27}\\
& = 3.98 \times 10^{-30}\mathrm{\ kg}\\
\end{aligned}
\)

Must mention strong nuclear force

Use of data

Sum of particle masses

Correct calculation of mass difference

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3.8.1.5

04.2

\(
\begin{aligned}
\text{Binding energy}& = mc^{2}\\
& = 3.98 \times 10^{-30} \times (3.00 \times 10^{8})^{2}\\
& = 3.58 \times 10^{-13}\mathrm{\ J}\\
& = \frac{3.569 \times 10^{-13}}{1.6 \times 10^{-13}}\\
& = 2.24 \mathrm{\ MeV}
\end{aligned}
\)

Answer in J

Answer in MeV

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3.8.1.5

04.3

The binding energy is 3 and 4 times bigger

The tritium nucleus has an additional neutron which is attracted to both the other two particles in the nucleus

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3.8.1.5

04.4

The tritium contains one proton and two neutrons, whereas the helium-3 contains one neutron and two protons

The strong nuclear force is charge independent, so the energy required to separate the nucleons is the same in each nucleus

The two protons in the helium-3 nucleus repel each other, so the energy needed to separate them is smaller

Comparison of the nuclei in terms of protons and neutrons

Recognition that effect of strong nuclear force is the same

Use of Coulomb repulsion

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3.8.1.5

05.1

A nuclear chain reaction happens when the neutrons from one reaction trigger subsequent (fission) reactions, creating a chain

This will be self-sustaining if the neutrons do not escape, or if the nuclei are sufficiently close, so this will not happen if the mass of fissionable nuclei is too small

Explanation of chain reaction

Condition for not self-sustaining

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3.8.1.7

05.2

Suggestion, for example:

The reactor did not produce enough radiation/was operating for long enough for shielding or cooling to be needed

Credit any sensible suggestion

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3.8.1.8

05.3

Suggestion, for example:

When a uranium nucleus undergoes fission, energy is released in the form of the kinetic energy of the fission fragments

This produces a temperature rise in the coolant, that drives a turbine and a generator

The fuel in the Chicago Pile-1 did not contain sufficient numbers of uranium nuclei for sufficient numbers of fission reactions to occur per second to produce a similar (or half of the) power.

Description of energy released in fission reaction

Description of link to power

Sensible suggestion for lack of energy per second

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3.8.1.8

05.4

They absorb thermal neutrons so can be used to shut down the reactor by being lowered into the reactor

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3.8.1.7

06.1

Momentum/energy was not conserved

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3.2.1.2

06.2

W+, W bosons

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3.2.1.4

06.3

The half-life is (relatively) short, and the nucleus emits beta radiation, which is not very ionising, so this would be an acceptable risk if the level of sodium was potentially responsible for a serious illness

Appreciation of the balance of risk and benefit

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3.8.1.3

06.4

When the neutron is absorbed, the atomic mass increases by one, atomic number does not change

When the beta particle is emitted, the atomic mass does not change, and the atomic number increases by one

Each change correct

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3.8.1.4

06.5

The majority of the decays leave the nucleus in an excited state with 4.14 MeV

The nucleus can decay from this level to 1.38 MeV, emitting a gamma ray of 4.14 – 1.38 MeV = 2.76 MeV

It then emits a gamma ray of energy 1.38 MeV, and decays to the ground state

Both energy level correct

Justification

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3.8.1.5

06.6

Exposure to a large number of neutrons could mean that fission reactions, which produce neutrons, may be too numerous; control measures such as control rods are not working

Geiger counters detect highly ionising radiation, but do not detect neutrons (unless modified)

Appropriate reason

Justification involving lack of ionisation power of neutrons

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3.8.1.7

07.1

A chain reaction happens when the neutrons produced in a fission reaction start subsequent reactions

Which usually happens when a moderator slows the neutrons down

The water acts as a moderator, slowing down neutrons so that they can be absorbed by uranium nuclei

Allow U isotope is stable/can not absorb neutrons

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3.8.1.7

07.2

Fissile means that the nucleus can split, usually when it absorbs a thermal neutron, to produce fission products and more neutrons, but non-fissile means that it cannot

Both needed for the mark

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3.8.1.6

07.3

Before: atomic mass = 235 + 1, atomic number = 92

After: atomic mass = 142 + 3, atomic number = 60

Atomic mass = 236 – 145 = 91, atomic number = 32

Atomic mass correct

Atomic number correct

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3.8.1.6

07.4

Energy per reaction = 150 × 1.6 × 10−13 J = 2.4 × 10−11 J

Power = 100 kW = 105 J s–1

\(
\begin{aligned}
\text{Decays per second}& =\frac{\text{power}}{\text{energy per reaction}} = \frac{10^5\mathrm{\ J\ s}^{-1}}{2.4 \times 10^{-11}\mathrm{\ J}}\\
& = 4.2 \times 10^{15}\text{ decays second}^{-1}
\end{aligned}
\)

Use of power

Explicit or implied

Answer

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3.4.1.7

07.5

Energy required to raise temperature of water and change state of water = \(mC\Delta \theta + mL\)

In one second: m [(4200 × 80) + (2 260 000)] = 100 000 J

\(
\begin{aligned}
& \text{Mass per second } =\frac{100\,000}{2\,596\,000}= 3.85 \times 10^{-2}\mathrm{\ kg s}^{-1}\\
& \text{Volume per second } = \frac{\text{mass per second}}{\text{density}} = \frac{0.0385}{0.6}= 6.4(2) \times 10^{-2}\mathrm{\ m}^3\mathrm{\ s}^{-1}
\end{aligned}
\)

Use of SHC and SLH

Correct mass per second

Correct volume per second

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3.6.2.1

08.1

Induced fission requires the absorption of a thermal/slow-moving neutron but spontaneous fission does not

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3.8.1.7

08.2

Atomic mass increases by one: 239

Atomic number increases by two: 94

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3.8.1.4

08.3

One from:

  • Plutonium has a higher probability of fission, so less fuel would be needed to produce the same energy as the energy per fission is about the same
  • The half-life is less, so would become less radioactive more quickly
  • There are fewer spontaneous fissions per kilogram, so there is less chance of the mass going critical/exploding

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3.8.1.6

08.4

Suggestion such as:

Electrons are moving faster at one end of the wire and move towards the lower temperature end and moving charge = current

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3.5.1.1

08.5

Plutonium-238 decays by alpha decay and it is easy to shield the patient from the radiation using a thin sheet of metal

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3.8.1.2

08.6

\(
\begin{aligned}
\text{Power} & = IV\\
& = 2 \times 10^{-3}\text{ A }\times 1.5\text{ V }= 3 \times 10^{-3}\text{\ W}\\
\end{aligned}
\) \(
\begin{aligned}
\text{Power density} &= 0.57\textrm{ W g}^{-1}\\
\text{Mass required} & = \frac{0.57}{3 \times 10^{-3}\mathrm{\ W}} = 190\mathrm{\ g}\\
\end{aligned}
\)

Reason: only pulses produced/continuous supply not needed

Power required

Answer

Reason

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3.5.1.4

Skills box answers

Question

Answer

1

Order of magnitude values: \(E_k \approx 10^{-14}\) J; \(q \approx 10^{-19}\) C; \(Q \approx 10^{-19}\) C; \(4 \pi \varepsilon_0 \approx 10^{-10}\), Fm-1, so \(r \approx 10^{-14}\) m.

2

Diameter of atom is \(\approx 10^{-10}\ m\), thickness of foil is \(0.5 \times 10^{-6}\mathrm{\ m } \approx 10^{-6}\). Number of atoms in the depth of foil = \(\frac{10^{-6}}{10^{-10}} = 10^4.\)

3

Gold is an extremely malleable material and can be beaten into very thin sheets without cracking.

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