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Physics
Binding energy of a Nitrogen nucleus [ 714 N ], given m [ 714 N ]=14.00307 u
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Q. Binding energy of a Nitrogen nucleus $\left[{ }_{7}^{14} N \right]$, given $m \left[{ }_{7}^{14} N \right]=14.00307\, u$
KCET
KCET 2022
Nuclei
A
206.5 MeV
21%
B
78 MeV
23%
C
104.7 MeV
46%
D
85 MeV
10%
Solution:
$BE =\left[ Zm _{ p }+( A - Z ) m _{ n }- m _{ X }\right] \times 931.5$
$Be =[7.05481+7.06069-14.0030] \times 931.5$
$=104.7\, MeV$