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Calculate the average binding energy per nucleon of `._(41)^(93)Nb` having mass 9.2.906 u.. |
Answer» In order to compute binding energy, let us first find the total mass of all protons and neutrosn in Nb and substarct mass of the Nb: Given: `m_(p) =1.007276 u` and `m_(n)=1.008665 u` Number of protons: `N_(p) =41` Number of neutrons: `N_(n) =93 -41=52` mass differences: `Delta m=41 m_(p) + 52 m_(n) -m_(Nb)` `=41(1.007825 u) +52(1.008665 u) -(92.9063768 u)` `=0.865028 u ` Thus, binding energy per nuclear is . `(E_(b))/(A)=((Delta)c^(2))/(A)=((0.865028u)(931.5MeV//u))/(93)` `=8. 66 MeV ` nucleon^(-1)`. |
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