If M_0 is the mass of isotope { }_{5}^{12} B, | vedclass

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Question

(A) The nuclear binding energy $(B.E.)$ is defined as the energy equivalent of the mass defect $(\Delta m)$.The isotope ${ }_{5}^{12} B$ has $Z = 5$ p

Vedclass · Accepted Answer

$(5 M_p + 7 M_n - M_0) C^2$

Vedclass · Answer

(A) The nuclear binding energy $(B.E.)$ is defined as the energy equivalent of the mass defect $(\Delta m)$.The isotope ${ }_{5}^{12} B$ has $Z = 5$ protons and $A - Z = 12 - 5 = 7$ neutrons.The mass defect $\Delta m$ is given by the difference between the sum of the masses of individual nucleons and the actual mass of the nucleus $(M_0)$:$\Delta m = (Z M_p + (A - Z) M_n) - M_0$$\Delta m = (5 M_p + 7 M_n - M_0)$Using Einstein's mass-energy equivalence relation,$B.E. = \Delta m C^2$:$B.E. = (5 M_p + 7 M_n - M_0) C^2$.

If $M_0$ is the mass of isotope ${ }_{5}^{12} B$,$M_p$ and $M_n$ are the masses of a proton and a neutron respectively,then the nuclear binding energy of the isotope is:

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