Question

Problem 3:

 If the mass of 3Li7 is 7.01653 amu, then find out binding energy per nucleon for 3Li7

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Answer 1

Answer:

E = ΔE / A

= Δm × 931 / A MeV  

Δm = (3mp + 4mn) – mass of Li7  

= (3 × 1.00759 + 4 × 1.008898) – 7.01653

= 0.04216

ΔE = 0.04216 × 931 / 7 = 39.25 / 7

= 5.6 MeV

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