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Nuclear Fission

The mass of ...

Question

The mass of $_{3} L i^{7}$ nucleus is $0.042$ a.m.u. less than the sum of masses of its nucleons. Find the binding energy per nucleon.

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Solution

We have Mass number = 7 and Atomic number= 3

If mass, $m = 1 u$ and coulomb force $c = 3$ X $10^{8} m s^{- 1}$ , then

$E = 931 M e V$ that is, $1 u = 931 M e V$

Binding Energy = $0.042$ X $931 = 39.10 M e V$

Therefore, BInding Energy per Nucleon $= \frac{B i n d i n g E n e r g y}{M a s s N u m b e r} = \frac{39.10}{7} = 5.58 = 5.6 M e V$

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