Respuesta :
The Binding Energy per Nucleon of an Al -27 nucleus Electron= 26.981541 amu, Proton = 1.007825 amu, Neutron = 1.008665 amu is 8.33 MeV
By dividing the total binding energy of the nucleus by the total number of nucleons, one can calculate the binding energy per nucleon. Mathematically,
Binding energy per nucleon is equal to the sum of all the nucleons.
we know the binding energy can be calculate with formulla:
E= Δmc²
Total binding energy (E) is 931.5 MeV/amu
first we should calculate mass defect (Δm)
Total binding energy (E) is 931.5 MeV/amu
proton = 1.007825 amu
Number of protons in Al-27 = 13
So, the total mass of protons ( mp) = 13.101725 amu
neutron = 1.008665 amu
Number of neutrons in Al-27 = 14
So, the total mass of total neutrons ( mn)= 14.121310 amu
Atomic mass of Al-27 26.981541 amu
Thus, the mass defect can be calculate as follows,
Δm=mp+mn−A
Δm= 0.241492 amu
so the binding energy,
E= Δm X total binding energy
E = 0.241494 amu x 931.5 MeV/amu
E =224.95 MeV
The Binding Energy per Nucleon can be calculate as follows:
Binding Energy per Nucleon = Binding Energy
Total number of nucleons
=224.95 MeV/27
= 8.33 MeV
To learn more about binding energy per nucleon please click on the given link: brainly.com/question/28305588
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