Calculate the binding energy per nucleon of an Al-27 nucleus given the following masses: ${ }_{13}^{27} \mathrm{Al}$ : 26.981541 amu; : ${ }_{1}^{1} \mathrm{H}$ 1.007838 amu; : ${ }_{0}^{1} \mathrm{n}$ 1.008665 amu.

Question

Quizplus · Accepted Answer

The binding energy can be calculated using the formula:
BE = [Z(mass of proton) + (A-Z)(mass of neutron) - mass of nucleus]c^2
where:
Z = number of protons = 13
A = mass number = 27
c = speed of light = 2.998 x 10^8 m/s

Substituting the given values, we get:
BE = [13(1.007838) + 14(1.008665) - 26.981541]c^2
BE = (7.288067) x 1.602 x 10^-13 J
BE/nucleon = BE/A = (7.288067 x 1.602 x 10^-13)/27 J/nucleon
BE/nucleon = 1.34 x 10^10 J/nucleon or 1.34 MeV/nucleon

Therefore, the binding energy per nucleon of an Al-27 nucleus is 1.34 MeV/nucleon, which corresponds to option B.

Calculate the Binding Energy Per Nucleon of an Al-27 Nucleus 1327Al{ }_{13}^{27} \mathrm{Al}1327​Al

Calculate the Binding Energy Per Nucleon of an Al-27 Nucleus ${ }_{13}^{27} \mathrm{Al}$