The mass of a nucleus _Z^AX is denoted by M(A | vedclass

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(A) The binding energy $(BE)$ of a nucleus is defined as the energy equivalent of the mass defect $(\Delta m)$.Mass defect is the difference between t

Vedclass · Accepted Answer

$BE = [ZM_p + (A - Z)M_n - M(A, Z)]c^2$

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(A) The binding energy $(BE)$ of a nucleus is defined as the energy equivalent of the mass defect $(\Delta m)$.Mass defect is the difference between the sum of the masses of individual nucleons and the actual mass of the nucleus.Number of protons = $Z$.Number of neutrons = $A - Z$.Mass of nucleons = $ZM_p + (A - Z)M_n$.Mass defect $\Delta m = [ZM_p + (A - Z)M_n - M(A, Z)]$.Using Einstein's mass-energy equivalence relation, $E = mc^2$, the binding energy is $BE = \Delta m c^2$.Therefore, $BE = [ZM_p + (A - Z)M_n - M(A, Z)]c^2$.

The mass of a nucleus $_Z^AX$ is denoted by $M(A, Z)$. If $M_p$ and $M_n$ are the masses of a proton and a neutron respectively, the binding energy of this nucleus is given by:

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The binding energy per nucleon of a nucleus ${}_Z X^A$ at rest is $6 \ MeV$. It undergoes $\beta^-$ decay as shown below:
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