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Date	November 2020	Marks available	1	Reference code	20N.2.HL.TZ0.6
Level	Higher level	Paper	Paper 2	Time zone	0 - no time zone
Command term	Suggest	Question number	6	Adapted from	N/A

Question

One possible fission reaction of uranium-235 (U-235) is

${}_{92}^{235}U + {}_{0}^{1}n \to {}_{54}^{140}{Xe} + {}_{38}^{94}{Sr} + 2 {}_{0}^{1}n$

Mass of one atom of U-235 $= 235 u$
Binding energy per nucleon for U-235 $= 7.59 MeV$
Binding energy per nucleon for Xe-140 $= 8.29 MeV$
Binding energy per nucleon for Sr-94 $= 8.59 MeV$

A nuclear power station uses U-235 as fuel. Assume that every fission reaction of U-235 gives rise to $180 MeV$ of energy.

A sample of waste produced by the reactor contains $1.0 kg$ of strontium-94 (Sr-94). Sr-94 is radioactive and undergoes beta-minus ( $β^{-}$ ) decay into a daughter nuclide X. The reaction for this decay is

${}_{38}^{94}{Sr} \to X + {\bar{v}}_{e} + e$ .

The graph shows the variation with time of the mass of Sr-94 remaining in the sample.

State what is meant by binding energy of a nucleus.

[1]

a(i).

Outline why quantities such as atomic mass and nuclear binding energy are often expressed in non-SI units.

[1]

a(ii).

Show that the energy released in the reaction is about $180 MeV$ .

[1]

a(iii).

Estimate, in $J {kg}^{- 1}$ , the specific energy of U-235.

[2]

b(i).

The power station has a useful power output of $1.2 GW$ and an efficiency of $36 %$ . Determine the mass of U-235 that undergoes fission in one day.

[2]

b(ii).

The specific energy of fossil fuel is typically $30 MJ {kg}^{- 1}$ . Suggest, with reference to your answer to (b)(i), one advantage of U-235 compared with fossil fuels in a power station.

[1]

b(iii).

Write down the proton number of nuclide X.

[1]

c(i).

State the half-life of Sr-94.

[1]

c(ii).

Calculate the mass of Sr-94 remaining in the sample after $10$ minutes.

[2]

c(iii).

Markscheme

energy required to «completely» separate the nucleons
OR
energy released when a nucleus is formed from its constituent nucleons ✓

Allow protons AND neutrons.

a(i).

the values «in SI units» would be very small ✓

a(ii).

$140 \times 8.29 + 94 \times 8.59 - 235 \times 7.59$ OR $184 « MeV »$ ✓

a(iii).

see $« energy = » 180 \times 10^{6} \times 1.60 \times 10^{- 19}$ AND $« mass = » 235 \times 1.66 \times 10^{- 27}$ ✓

$7.4 \times 10^{13} « J {kg}^{- 1} »$ ✓

b(i).

energy produced in one day $= \frac{1.2 \times 10^{9} \times 24 \times 3600}{0.36} = 2.9 \times 10^{14} « J »$ ✓

mass $= \frac{2.9 \times 10^{14}}{7.4 \times 10^{13}} = 3.9 « kg »$ ✓

b(ii).

«specific energy of uranium is much greater than that of coal, hence» more energy can be produced from the same mass of fuel / per $kg$
OR
less fuel can be used to create the same amount of energy ✓

b(iii).

$39$ ✓

Do not allow ${}_{39}^{94}X$ unless the proton number is indicated.

c(i).

$75 « s »$ ✓

c(ii).

ALTERNATIVE 1

$10 \min$ $= 8 t_{1 / 2}$ ✓

mass remaining $= 1.0 \times {(\frac{1}{2})}^{8} = 3.9 \times 10^{- 3} « kg »$ ✓

ALTERNATIVE 2

decay constant $= « \frac{\ln 2}{75} = » 9.24 \times 10^{- 3} « s^{- 1} »$ ✓

mass remaining $= 1.0 \times e^{- 9.24 \times 10^{- 3} \times 600} = 3.9 \times 10^{- 3} « kg »$ ✓

c(iii).

Examiners report

Generally, well answered but candidates did miss the mark by discussing the constituents of a nucleus rather than the nucleons, or protons and neutrons. There seemed to be fewer comments than usual about 'the energy required to bind the nucleus together'.

a(i).

Well answered with some candidates describing the values as too large or small.

a(ii).

Well answered.

a(iii).

This caused problems for some with mass often correctly calculated but energy causing more difficulty with the eV conversion either being inaccurate or omitted. Candidates were allowed error carried forward for the second mark as long as they were dividing an energy by a mass.

b(i).

Most candidates had the right idea, but common problems included forgetting the efficiency or not converting to days.

b(ii).

HL only. This was well answered.

b(iii).

Most candidates answered this correctly.

c(i).

Most candidates answered this correctly.

c(ii).

This was answered well with most candidates (even at HL) going down the number of half-lives route rather than the exponential calculation route.

c(iii).