Calculating Energy Released in Nuclear Reactions (CIE A Level Physics)

When uranium-235 nuclei undergo fission by absorbing slow-moving neutrons, two reactions are possible:

$\mathrm{Reaction} 1: {}_{92}{}^{235}\mathrm{U} + {}_0{}^1\mathrm{n} \to {}_{54}{}^{139}\mathrm{Xe} + {}_{38}{}^{95}\mathrm{Sr} + 2{}_0{}^1\mathrm{n} + \mathrm{energy}$

$\mathrm{Reaction} 2: {}_{92}{}^{235}\mathrm{U} + {}_0{}^1\mathrm{n} \to 2{}_{46}{}^{116}\mathrm{Pd} + x\mathrm{c} + \mathrm{energy}$

Answer:

(a)

Step 1:  Balance the number of protons on each side (bottom number)

92 = (2 × 46) + xn_p (where n_p is the number of protons in c)

xn_p = 92 – 92 = 0

Therefore, c must be a neutron

Step 2: Balance the number of nucleons on each side

235 + 1 = (2 × 116) + x

x = 235 + 1 – 232 = 4

Therefore, 4 neutrons are generated in the reaction

(b)

Step 1: Find the binding energy of each nucleus

Total binding energy of each nucleus = Binding energy per nucleon × Mass number

Binding energy of ⁹⁵Sr = 8.74 × 95 = 830.3 MeV

Binding energy of ¹³⁹Xe = 8.39 × 139 = 1166.21 MeV

Binding energy of ²³⁵U = 7.60 × 235 = 1786 MeV

Step 2: Calculate the difference in energy between the products and reactants

Energy released in reaction 1 = E_Sr + E_Xe – E_U

Energy released in reaction 1 = 830.3 + 1166.21 – 1786

Energy released in reaction 1 = 210.5 MeV

(c)

• Since reaction 1 releases more energy than reaction 2, its end products will have a higher binding energy per nucleon
  • Hence they will be more stable

• This is because the more energy is released, the further it moves up the graph of binding energy per nucleon against nucleon number (A)
  • Since at high values of A, binding energy per nucleon gradually decreases with A

• Nuclear reactions will tend to favour the more stable route, therefore, reaction 1 is more likely to happen