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Calculate the binding energy per nucleon...

Calculate the binding energy per nucleon of `._20Ca^(40)` nucleus. Given m `._20Ca^(40)=39.962589u` , `M_(p) =1.007825u` and `M_(n)=1.008665 u` and Take `1u=931MeV`.

A

`E_(bn)=5.317` Me V/nucleon

B

`E_(bn)=6.151` Me V/nucleon

C

`E_(bn)=4.236` Me V/nucleon

D

`E_(bn)=8.547` Me V/nucleon

Text Solution

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The correct Answer is:
D
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