9702 · 23.1

Mass defect and nuclear binding energy — practice questions

Practice and worked examples for 9702 Mass defect and nuclear binding energy. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

A nucleus of Helium-4 ( $^4_2\text{He}$ ) has an actual mass of $4.001506 \text{ u}$ . Given the mass of a proton is $1.007276 \text{ u}$ and a neutron is $1.008665 \text{ u}$ , calculate the mass defect and the binding energy of Helium-4 in MeV.

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Identify components: Helium-4 has 2 protons (Z=2) and 2 neutrons (N=2).
Calculate total mass of individual nucleons: $M_{individual} = (2 \times m_p) + (2 \times m_n)$ $M_{individual} = (2 \times 1.007276 \text{ u}) + (2 \times 1.008665 \text{ u})$ $M_{individual} = 2.014552 \text{ u} + 2.017330 \text{ u} = 4.031882 \text{ u}$
Calculate mass defect ( $\Delta m$ ): $\Delta m = M_{individual} - M_{nucleus}$ $\Delta m = 4.031882 \text{ u} - 4.001506 \text{ u} = 0.030376 \text{ u}$
Convert mass defect to binding energy (MeV): Using the conversion 1 u $\approx$ 931.5 MeV: Binding Energy = $0.030376 \text{ u} \times 931.5 \text{ MeV/u}$ Binding Energy $\approx 28.295 \text{ MeV}$ (Rounded to 3 significant figures, Binding Energy = $28.3 \text{ MeV}$ )

Worked example 2

Consider the fission of a Uranium-235 nucleus after it absorbs a slow neutron. One possible reaction is: $^1_0n + ^{235}_{92}U \rightarrow ^{141}_{56}Ba + ^{92}_{36}Kr + 3(^1_0n)$ Calculate the energy released in this reaction in MeV. Use the following data:

Mass of a neutron ( $^1_0n$ ) = 1.008665 u
Mass of a $^{235}_{92}U$ nucleus = 235.043930 u
Mass of a $^{141}_{56}Ba$ nucleus = 140.914411 u
Mass of a $^{92}_{36}Kr$ nucleus = 91.926156 u
1 u is equivalent to 931.5 MeV.

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Calculate the total mass of the reactants (before fission): Mass_reactants = Mass( $^{235}U$ ) + Mass( $^1n$ ) Mass_reactants = 235.043930 u + 1.008665 u = 236.052595 u
Calculate the total mass of the products (after fission): Mass_products = Mass( $^{141}Ba$ ) + Mass( $^{92}Kr$ ) + 3 × Mass( $^1n$ ) Mass_products = 140.914411 u + 91.926156 u + 3 × (1.008665 u) Mass_products = 140.914411 u + 91.926156 u + 3.025995 u Mass_products = 235.866562 u
Calculate the mass defect (Δm): The mass defect is the difference between the initial and final mass. Δm = Mass_reactants - Mass_products Δm = 236.052595 u - 235.866562 u = 0.186033 u
Convert mass defect to energy released (E): Energy is released because the mass has decreased. E = Δm × 931.5 MeV/u E = 0.186033 u × 931.5 MeV/u E ≈ 173.28 MeV The energy released is approximately 173 MeV.

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