Question

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

• A. \((5M_p + 7M_n - M_0)C^2\)
• B. \((M_0 - 5M_p)C^2\)
• C. \((M_0 - 12M_n)C^2\)
• D. \((M_0 - 5M_p - 7M_n)C^2\)

Answer 1

The Correct Option is A

The binding energy (\(B.E.\)) of a nucleus is given by:
\[B.E. = \Delta m c^2,\]
where \(\Delta m\) is the mass defect.
The mass defect for the isotope \({}^{12}_5 B\) is:
\[\Delta m = (5M_p + 7M_n) - M_0.\]
Substituting \(\Delta m\) into the binding energy equation:
\[B.E. = (5M_p + 7M_n - M_0)c^2.\]
Thus, the nuclear binding energy of the isotope is:
\[B.E. = (5M_p + 7M_n - M_0)c^2.\]