First slide

Nuclear Binding energy

Question

The binding energies per nucleon are 5.3 MeV, 6.2 MeV and 7.4 MeV for the nuclei with mass numbers 3, 4 and 5 respectively. If one nucleus of mass number 3 combines with one nucleus of mass number 5 to give two nuclei with mass number 4. Then

Moderate

Solution

$\begin{array}{l} B_{i} = (3 \times 5 .3 + 5 \times 7 .4) MeV = 52 .9 MeV \\ B_{f} = (4 \times 6 .2) \times 2 MeV = 49 .6 MeV \end{array}$
Since $B_{i} > B_{f},$ energy is absorbed
$∴$ Energy absorbed = $B_{i} - B_{f} = 3 .3 MeV$

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