1.

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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