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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`. |
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Answer» In a nucleus of `._20Ca^(40)`, number of protons=20, number of neutrons `=40-20=20` Total mass of 20 protons and 20 neutrons `=20m_p+20m_n=20(m_p+m_n)` `=20(1.007825+1.008665)=40.3298u` Mass defect, `Delta m=40.3298-39.962589` `=0.367211 u` Total `B.E=0.367211xx931MeV` `=341.873441 MeV` `BE//"nucleon"=341.873441/40=8.547MeV//N` |
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