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| 1. |
Obtain the binding energy of a nitrogen nucleus 147N Given m = 147N 14.00307 u. |
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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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