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:
- \((5M_p + 7M_n - M_0)C^2\)
- \((M_0 - 5M_p)C^2\)
- \((M_0 - 12M_n)C^2\)
- \((M_0 - 5M_p - 7M_n)C^2\)
The Correct Option is A
Approach Solution - 1
To find the nuclear binding energy of the isotope ^{12}_{5}B, we need to first understand the components involved. The nuclear binding energy is the energy required to separate an atomic nucleus into its individual protons and neutrons. It can be calculated using the mass defect, which is the difference between the sum of the masses of the individual nucleons and the actual mass of the nucleus.
- We have the isotope \(^{{12}}_{{5}}B\), which means it consists of 5 protons and \(12 - 5 = 7\) neutrons.
- The total mass of the separated nucleons would be \(5M_p + 7M_n\), where \(M_p\) is the mass of a proton, and \(M_n\) is the mass of a neutron.
- The mass defect \(\Delta m\) is then given by the difference: \(\Delta m = (5M_p + 7M_n) - M_0\) where \(M_0\) is the mass of the isotope nucleus.
- The nuclear binding energy \(E_b\) is found using Einstein's mass-energy equivalence principle, \(E = \Delta mc^2\), where \(c\) is the speed of light.
- Thus, substituting the expression for mass defect into the equation, the binding energy \(E_b\) becomes: \(E_b = (5M_p + 7M_n - M_0)C^2\)
Therefore, the correct option that represents the nuclear binding energy of the isotope \(^{{12}}_{{5}}B\) is:
\((5M_p + 7M_n - M_0)C^2\)
Approach Solution -2
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:
\[B.E. = (5M_p + 7M_n - M_0)c^2.\]
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