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Binding Energy

Obtain the bi...

Question

Obtain the binding energy (in $M e V$ ) of a nitrogen nucleus $(_{7}^{14} N)$ , given $m (_{7}^{14} N) = 14.00307 u$

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Solution

$m_{p} = 1.007825 a m u; m_{n} = 1.008665 a m u$
The nitrogen nucleus contains $7$ protons and $7$ neutrons
The binding energy of nitrogen nucleus
$= [{7 \times 1.007825 + 7 \times 1.008665} - 14.00307] \times 931.5 = [{7.054775 + 7.060655} - 14.00307] \times 931.5 = 104.66 M e V$

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