The binding energy per nucleon for $_{6} C^{12}$ nucleons is $7.68$ MeV and that for $_{6} C^{13}$ is $7.47$ MeV. Calculate the energy required to remove a neutron from $_{6} C^{13}$ nucleons.

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Solution

Given data:
Binding energy per nucleon of
$_{6} C^{13} = 7.47$ MeV
Binding energy per nucleon of $_{6} C^{12} = 7.68$ MeV
Binding energy of neutron $= ?$
Solution:
We can write the reaction as:
$_{6} C^{13} \to_{6} C^{12} +_{0} n^{1}$
Total binding energy of $_{6} C^{13} = 7.47 \times 13 = 97.11$ MeV
Total binding energy of $_{6} C^{12} = 7.68 \times 12 = 92.16$ MeV
Total binding energy of the reactant $=$ Total binding energy of the product
$∴ 97.11$ MeV $= 92.16$ MeV $+$ Binding energy of a neutron.
$∴$ Binding energy of a neutron $= 97.11 - 92.16$
$= 4.95$ MeV.

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