Question
Both fission and fusion reactions are potential sources of nuclear energy.
(a) Compare and contrast the nuclear changes and products formed in these processes
giving one similarity and one difference.
Similarity:
Difference:
(b) Uranium is converted into a gaseous compound for enrichment.
(i) Identify the gaseous compound.
(ii) Determine the percentage difference in the rate of diffusion of molecules containing \(^{235}U\) compared to those containing \(^{238}U\). The molar masses of these molecules are 349 and 352 respectively. Use section 1 of the data booklet.
(iii) (iii) Show how the dependence of the rate of diffusion on molar mass arises from kinetic theory. Use section 1 of the data booklet and:
\(E = \frac{1}{2} mv^2\),
where E is energy of the particle, m its mass and v its velocity.
(c) Some reactors convert \(^{238}U\) into another nucleus that can also undergo fission.
(i) Complete the equation for this process by identifying the reacting particle, X, and the isotope formed, Y.
\(^{238}U + X \rightarrow ^{239}U \rightarrow Y + 2^o \beta \)
X:
Y:
(ii) The intermediate, \(^{239}U\), has a half-life of 23 minutes. Outline what is meant by half-life.
Answer/Explanation
Answer:
(a) Similarity:
increase binding energy «per nucleon»
OR
«can» produce chain reactions
Difference:
fusion forms one product/products with a greater «atomic» mass AND fission
multiple products/products with a lower mass
OR
fission produces «long lived» radioactive products/nuclear waste AND fusion
does not
(b) (i) uranium hexafluoride/\(UF_6\)
(ii) «\(^{238}U\) rate = \(^{235}U\) rate x \((349/352)^{1⁄2}\) = \(^{235}U\) rate x » 0.996
OR
«\(^{235}U\) rate = \(^{238}U\) rate x \((352/349)^{1⁄2}\) = \(^{238}U\) rate x » 1.004
percentage difference «= (1-0.996) x 100» = 0.4%
(iii) rate of diffusion proportional to v
v inversely proportional to √m
OR
v = \((2E/m)^{1⁄2}\)
OR
\(\frac{v_1}{v_2} = \frac{(2E/m_1)^{1/2}}{(2E/m_2)^{1/2}}\)
(c) (i) X: «1»n/neutron
Y: \(^{239}Pu\)/Pu – 239/plutonium-239
(ii) time for half the number of atoms/nuclei/mass to decay
Question
$1.57 \% \text { of the mass of a rock weighing } 46.5 \mathrm{~kg} \text { is uranium(IV) oxide, } \mathrm{UO}_2 .99 .28 \% \text { of the uranium atoms in the rock are uranium-238, }{ }^{238} U$
a. Show that the mass of the ${ }^{238} \mathrm{U}$ isotope in the rock is $0.639 \mathrm{~kg}$.
b. The half-life of ${ }^{238} \mathrm{U}$ is $4.46 \times 10^9$ years. Calculate the mass of ${ }^{238} \mathrm{U}$ that remains after $0.639 \mathrm{~kg}$ has decayed for $2.23 \times 10^{10}$ years.
c. Outline a health risk produced by exposure to radioactive decay.
d. Deduce the nuclear equation for the decay of uranium-238 to thorium-234.
e. Thorium-234 has a higher binding energy per nucleon than uranium-238. Outline what is meant by the binding energy of a nucleus.
\text { f. Determine the nuclear binding energy, in } \mathrm{J} \text {, of } { }^{238} \mathrm{U} \text { using sections } 2 \text { and } 4 \text { of the data booklet. }The mass of the ${ }^{238} \mathrm{U}$ nucleus is $238.050786 \mathrm{amu}$.
▶️Answer/Explanation
Markscheme
a. ” $\frac{\text { mass } \%}{\text { fraction of } \mathrm{U} \text { in } \mathrm{UO}_2}=” \frac{238.03}{238.03+2 \times 16} / 0.881 / 88.1 \%$
$46.5 \ll \mathrm{kg} \gg \times 0.0157 \times 0.881 \times 0.9928 \ll=0.639 \mathrm{~kg} »$
Award [1 max] for omitting mass composition (giving $0.725 \mathrm{~kg}$ ).
M2 is for numerical setup, not for final value of $0.639 \mathrm{~kg}$.
b. Alternative 1
$« \frac{2.23 \times 10^{10} \text { year }}{4.46 \times 10^9 \text { year }}=» 5.00$ «half – lives $»$
$\ll m=0.639 \mathrm{~kg} \times(0.5)^5=» 0.0200 \ll \mathrm{kg} »$
Alternative 2
$$
\begin{aligned}
& \left\langle\lambda=\frac{\ln 2}{4.46 \times 10^9 \text { year }}=» 1.554 \times 10^{-10} \text { «year }^{-1} »\right. \\
& \ll m=0.639 \mathrm{~kg} \times e^{-1.554 \times 10^{-10} \text { year }^{-1} \times 2.23 \times 10^{10} \text { year }}=» 0.0200 \ll \mathrm{kg} » \\
&
\end{aligned}
$$
Award [2] for correct final answer.
c. Any one:
«genetic» mutations
«could cause” cancer $\checkmark$
Accept specific named types of cancer.
cells «in body» altered
cells «in body» cannot function
damaged DNA/proteins/enzymes/organs/tissue
«radiation» burns $\checkmark$
hair loss
damage in foetuses
damages/weakens immune system
d. $\mathrm{U}_{92}^{238} \rightarrow \mathrm{Th}_{90}^{234}+\mathrm{He}_2^4$
Do not penalize missing atomic numbers in the equation.
Accept “a.” for “He”.
e. energy required to separate a nucleus into protons and neutrons/nucleons
OR
energy released when nucleus was formed from «individual/free/isolated» protons and neutrons/nucleons
Do not accept “energy released when atom was formed”.
f. 238.050786 «amu» $\times 1.66 \times 10^{-27}\left\langle\mathrm{~kg} \mathrm{amu}{ }^{-1}\right.$ »
OR
$3.95 \times 10^{-25} \varkappa \mathrm{kg} »$
$\left(92 \times 1.672622 \times 10^{-27}\right)+\left(146 \times 1.674927 \times 10^{-27}\right)-3.95 \times 10^{-25}$
OR
$3.42 \times 10^{-27} / 3 \times 10^{-27}\langle\mathrm{~kg}\rangle$
$\left\langle E=m c^2=3.42 \times 10^{-27} \times\left(3.00 \times 10^8\right)^2=» 3.08 \times 10^{-10}\langle\mathrm{~J} 》\rangle\right.$
Accept answers in the range “2. $7 \times 10^{-10}-3.1 \times 10^{-10}$ «J””
Award [3] for correct final answer.