1.

Obtain the binding energy of a nitrogen nucleus 147N Given m = 147N 14.00307 u.

Answer»

Here Z = 7 and A = 14, A – Z = 14 – 7 = 7

∴ Mass defect

= [Z mH + (A – Z)mn – Mn] u

= (7 × 1.00783 + 7 × 1.00867 – 14.00307) u

= (7.05481 + 7.06069 - 14.00307) u = 0.11243 u

Since I u = 931.5 MeV

∴ B.E. of 14N = 0.11243 × 931 Mev

= 104.7 MeV.



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