If the mass of a proton is 1.007825 u and the mass of the neutron is 1.008665 u, the mass defect in the formation of \( {}^{7}_{3}\mathrm{Li} \) nucleus is: (take the mass of Lithium nucleus as 7.000000 u)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0.058135 u
The mass defect is the total mass of the separate nucleons minus the mass of the nucleus. \( {}^{7}_{3}\mathrm{Li} \) has \( Z = 3 \) protons and \( A - Z = 4 \) neutrons.
\( \displaystyle \Delta m = Z m_p + (A - Z) m_n - M_{\mathrm{nucleus}} \)
\( Z m_p = 3 \times 1.007825 = 3.023475\,\mathrm{u} \)
\( (A - Z) m_n = 4 \times 1.008665 = 4.034660\,\mathrm{u} \)
\( \displaystyle \Delta m = 3.023475 + 4.034660 - 7.000000 = 0.058135\,\mathrm{u} \)