Show that the mass of the ²³⁸U isotope in the rock is $0.639 \mathrm{kg}$.

The half-life of ²³⁸U is $4.46\times {10}^9$ years. Calculate the mass of ²³⁸U that remains after $0.639 \mathrm{kg}$ has decayed for $2.23\times {10}^{10}$ years.

Outline a health risk produced by exposure to radioactive decay.

Deduce the nuclear equation for the decay of uranium-238 to thorium-234.

Thorium-234 has a higher binding energy per nucleon than uranium-238. Outline what is meant by the binding energy of a nucleus.

Determine the nuclear binding energy, in $\mathrm{J}$, of ${}_{238}\mathrm{U}$ using sections 2 and 4 of the data booklet.

The mass of the ${}_{238}\mathrm{U}$ nucleus is $238.050786 \mathrm{amu}$.

$«\frac{\mathrm{mass} \%}{\mathrm{fraction} \mathrm{of} \mathrm{U} \mathrm{in} {\mathrm{UO}}_2}=»\frac{238.03}{238.03+2\times 16}$ /$0.881$/$88.1 \%$ ✔

$46.5«\mathrm{kg}»\times 0.0157\times 0.881\times 0.9928«=0.639 \mathrm{kg}»$ ✔

Award [1 max] for omitting mass composition (giving $\mathit{0}\mathit{.}\mathit{725}\mathit{ }kg$).

M2 is for numerical setup, not for final value of $\mathit{0}\mathit{.}\mathit{639}\mathit{ }kg$.

Alternative 1
$«\frac{2.23\times {10}^{10} \mathrm{year}}{4.46\times {10}^9 \mathrm{year}}=»5.00«\mathrm{half}-\mathrm{lives}»$ ✔

$«m=0.639 \mathrm{kg}\times (0.5{)}^5=»0.0200«\mathrm{kg}»$ ✔

Alternative 2
$«𝜆=\frac{\ln 2}{4.46\times {10}^9 \mathrm{year}}=»1.554\times {10}^{-10}«{\mathrm{year}}^{-1}»$ ✔

$«m=0.639 \mathrm{kg}\times {𝑒}^{–1.554\times {10}^{–10 {\mathrm{year}}^{–1}}\times 2.23\times {10}^{10 \mathrm{year}}}=»0.0200«\mathrm{kg}»$ ✔

Award [2] for correct final answer.

Any one:

«genetic» mutations ✔

«could cause» cancer ✔
Accept specific named types of cancer.

cells «in body» altered ✔

cells «in body» cannot function ✔

damaged DNA/proteins/enzymes/organs/tissue ✔

«radiation» burns ✔

hair loss ✔

damage in foetuses ✔

damages/weakens immune system ✔

${}_{92}{}^{238}\mathrm{U}\to {}_{90}{}^{234}\mathrm{Th}+{}_2{}^4\mathrm{He}$ ✔

Do not penalize missing atomic numbers in the equation.

Accept “$\alpha$” for "$He$”.

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”.

$238.050786«\mathrm{amu}»\times 1.66\times {10}^{-27}«\mathrm{kg} {\mathrm{amu}}^{-1}»$
OR

 ✔

$\left(92\times 1.672622\times {10}^{-27}\right)+\left(146\times 1.674927\times {10}^{-27}\right)-3.95\times {10}^{-25}$
OR
$3.42\times {10}^{-27}/3\times {10}^{-27}«\mathrm{kg}»$ ✔

$«E=m{c}^2=3.42\times {10}^{-27}\times (3.00\times {10}^8{)}^2=»3.08\times {10}^{-10}«\mathrm{J}»$ ✔

Accept answers in the range “$\mathit{2}\mathit{.}\mathit{7}\mathit{\times }{\mathit{10}}^{\mathit{-}\mathit{10}}\mathit{–}\mathit{3}\mathit{.}\mathit{1}\mathit{\times }{\mathit{10}}^{\mathit{-}\mathit{10}}\mathit{«}J\mathit{»}$”.

Award [3] for correct final answer.

This question was generally well answered. Many candidates approached this question using amount, in mol, and converting to mass at the end. Some candidates omitted the mass composition however, resulting in a mass of 0.725 kg, which yielded [1 max].

The question involving half-life was very well answered and most scored both marks giving a final answer of 0.0200 kg, via different methods of calculation.

This question on outlining a health risk produced by exposure to radioactive decay posed no difficulty and the most common, correct answers included "could cause cancer" and "can damage DNA".

Most scored the one mark for the nuclear equation for the decay of uranium-238 to thorium-234. The most common error was including a neutron on the product side

Although many scored the one mark for outlining what is meant by the binding energy of a nucleus, the question was often answered as energy involved in the formation of an atom. Some candidates did not understand the difference between nucleus and nucleons.

This demanding question on the determination of the nuclear binding energy was well executed and many scored all three marks. Even the weaker candidates still managed to gain an ECF mark for M3 by using the E = mc² equation. The general performance on this type of calculation was much better than in previous sessions.