*Question *

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

The following data are available.

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

(i) State what is meant by binding energy of a nucleus. [1]

(ii) Outline why quantities such as atomic mass and nuclear binding energy are often expressed in non-SI units. [1]

(iii) Show that the energy released in the reaction is about 180 MeV. [1]

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

(i) Estimate, in J kg^{–1}, the specific energy of U-235. [2]

(ii) 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]

(iii) 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]

**▶️Answer/Explanation**

Ans:

**a i**

energy required to «completely» separate the nucleons

OR

energy released when a nucleus is formed from its constituent nucleons ✓

**a ii** the values «in SI units» would be very small

**a iii **140 × 8.29 + 94 × 8.59 – 235 × 7.59 OR 184 «MeV»

**b i** see = « «energy=» 180 ×10^{6} × 1.60 ×10^{-19} AND « mass=» 235 × 1.66 × 10^{-27} , 7.4 × 10^{13} «J kg^{-1} »

**b ii**

energy produced in one day = \(\frac{1.2\times 109\times 22\times 3600}{0.36}\) = 2.9 ×10^{14} «J »

mass =\(\frac{2.9\times 10^{14}}{7.4\times 10^{15}}\)

**b iii**

«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 ✓

*Question*

This question is about binding energy and mass defect.

(i) The nuclear mass of the nuclide helium-3 \(\left( {_2^3{\text{He}}} \right)\) is 3.014931 u. Show that the binding energy per nucleon for the nuclide is about 2.6 MeV.

(ii) The binding energy per nucleon for deuterium \(\left( {_1^2{\text{H}}} \right)\) is 1.11 MeV. Calculate the energy change in the following reaction.

\[_1^2{\text{H}} + _1^1{\text{H}} \to _2^3{\text{HE}} + \gamma \]

(iii) The cross on the grid shows the binding energy per nucleon and nucleon number *A* of the nuclide nickel-62.

On the grid, sketch a graph to show how the average binding energy per nucleon varies with nucleon number *A*.

(iv) State and explain, with reference to your sketch graph, whether energy is released **or **absorbed in the reaction in (b)(ii).

**▶️Answer/Explanation**

## Markscheme

(i) mass defect \(\left( {[2 \times 1.007276 + 1.008665] – 3.014931 = } \right){\text{ }}0.008286{\text{ u}}\);

binding energy per nucleon \(\left( {\frac{{0.008286 \times 931.5}}{3} = } \right)\frac{{7.7}}{3}\)\(\,\,\,\)** or**\(\,\,\,\)\(2.58{\text{ (MeV)}}\);

(ii) binding energy of left-hand side \( = 1.11 \times 2\) and binding energy of right-hand side \( = 3 \times 2.6\); } *(both needed) (allow ECF)*

energy release \( = 5.58{\text{ (MeV)}}\); *(ignore sign)*

(iii) line goes through Ni point and nickel is the maximum ±2 small squares horizontally; } *(allow Fe-56 as maximum – this is just outside the range allowed)*

line starts at 0, downward trend for *A *after 62, trend after nickel less steep than before;

*Line must go through part of the X to award first marking point. *

*Line must not flatten out to award second marking point. *

*Allow smooth curve for low A. *

*Allow incorrect variations at low A.*

(iv) nucleus produced in the reaction is higher up the curve than the reactants / *OWTTE*; } *(must see reference to graph)*

reference to binding energy/other valid reason results in energy release;

*Award **[0] **for a bald correct answer.*

*Award **[0] **for any discussion of fission.*