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InterviewSolution
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Find the binding energy of a nucleus consisting of equal numbers of protons and neutrons and having the radius one and a half time smaller than of `Al^(27)` nucleus. |
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Answer» Correct Answer - `Be^(8),E_(b)=56.5MeV` The mass of defect is given by `E_(b)=Deltamc^(2)` where `Delta m =` mass defect If `m_(n)=` mass of neutron, `m_(p)=` mass of proton, `m_(N)=` mass of formed nucleus For binding energy of `._(z)X^(A)`, Here the number of proton `=Z`, The number of neutron `=A-Z` `:. Deltam =Zm_(p)+(A-Z)m_(n)-m_(N)` If atomic masses of the neucleus is given, then `:. Delta m=Zm_(p)+Zm_(e )+(A-Z)m_(n)-(m_(N)+Zm_(e ))` `=Z(m_(p)+m_(e ))+(A-Z)m_(n)-M_(N)` `=ZM_(H)+(A-Z)m_(n)-M_(N)` Here `M_(H)=` atomic mass of hydrogen `M_(N)=` atomic mass of formed atom. `:." "R=R_(0)A^(1//3)` `R_(Al)=R_(0)(27)^(1//3)` `R=R_(0)A^(1//3)` But `R=(R_(Al))/((3)/(2))=(2)/(3)R_(Al)` or `R_(0)A^(1//3)=(2)/(3)R_(0)(27)^(1//3)` `:. " " A=((2)/(3))^(3)=27=(8)/(27)xx27=8` From periodic table, the element is `Be^(8)` Atomic masses are `1.007825` for hydrogen, `1.008665` for nuetron, `8.00531` for `Be` `Deltam=(4m_(H)+4m_(n)-M_(N))"amu"` `=(4xx1.007825+4xx1.008665-8.00531)"amu"` `=(4.0313+4.034660-8.00531)"amu"` `E_(b)=Delta mc^(2)=0.06065xx931 MeV` `=56.4651 MeV=56.5 MeV` |
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