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Q. Binding energy of a Nitrogen nucleus $\left[{ }_{7}^{14} N \right]$, given $m \left[{ }_{7}^{14} N \right]=14.00307\, u$

KCETKCET 2022Nuclei

Solution:

$BE =\left[ Zm _{ p }+( A - Z ) m _{ n }- m _{ X }\right] \times 931.5$
$Be =[7.05481+7.06069-14.0030] \times 931.5$
$=104.7\, MeV$