Find the binding of the nucleus of lithium isotope $_{3} L i^{7}$ and hence find the binding energy per nucleon in it $(M_{_{3} L i^{7}} = 7.014353 amu, M_{_{1} H^{1}} = 1.007826$ , mass of neutron = $1.00867 u$ )

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Solution

Mass defect, $Δ m = 3 H + (7 - 3) n - L i = 3 H + 4 n - L i = 3 \times 1.007826 u + 4 \times 1.00867 u - 7.014353 u = 0.043805 u$

so binding energy $B E = Δ m \times 931.5 M e v = 0.043805 \times 931.5 M e v = 40.8 M e v$

So binding energy per nucleon will be $\frac{40.8 M e v}{mass number} = \frac{40.8}{7} = 5.829 M e v / n u c l e o n$

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_{3} L i^{7}

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