(i) Define binding energy of a nucleus.

(ii) The mass of a nucleus of plutonium \(\left( {_{94}^{239}{\rm{Pu}}} \right)\) is 238.990396 u. Deduce that the binding energy per nucleon for plutonium is 7.6 MeV.

The graph shows the variation with nucleon number A of the binding energy per nucleon.

Plutonium \(\left( {{}_{94}^{239}{\rm{Pu}}} \right)\) undergoes nuclear fission according to the reaction given below.

\[{}_{94}^{239}{\rm{Pu}} + {}_0^1{\rm{n}} \to {}_{38}^{91}{\rm{Sr}} + {}_{56}^{146}{\rm{Ba}} + x{}_0^1{\rm{n}}\]

(i) Calculate the number x of neutrons produced.

(ii) Use the graph to estimate the energy released in this reaction.

Stable nuclei with a mass number greater than about 20, contain more neutrons than protons. By reference to the properties of the nuclear force and of the electrostatic force, suggest an explanation for this observation.

(i) the (minimum) energy required to completely separate the nucleons of a nucleus / the energy released when a nucleus is assembled;

(ii) mass defect is 94×1.007276+145×1.008665−238.990396=1.95u;
binding energy is 1.95×931.5=1816MeV;
binding energy per nucleon is \(\frac{{1816}}{{239}}\)MeV;
=7.6MeV

(i) x=3;

(ii) binding energy of plutonium is 7.6×239=1816≈1800MeV
(known in (ii))
binding energy of products is 8.6×91+8.2×146=1980≈2000MeV;
energy released is (2000−1800)=200MeV;

the electric force is repulsive/tends to split the nucleus;
the electric force acts on protons, the strong nuclear force acts on nucleons;
the nuclear force is attractive/binds the nucleons;
but the electric force is long range whereas the nuclear force is short range;
so adding more neutrons (compared to protons) contributes to binding and does not add to tendency to split the nucleus / a proton repels every other proton (in the nucleus) so extra neutrons are needed for binding;