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 ${^7_3Li}$ nucleus is: (take the mass of Lithium nucleus as 7.000000 u)

0.058135 u

The correct answer is Option (1) → 0.058135 u

Given:

Mass of proton, $m_p = 1.007825\ \text{u}$

Mass of neutron, $m_n = 1.008665\ \text{u}$

Mass of $^7_3\text{Li}$ nucleus, $m_\text{Li} = 7.000000\ \text{u}$

Number of protons, $Z = 3$

Number of neutrons, $N = 7 - 3 = 4$

Mass defect, $\Delta m = [Z m_p + N m_n] - m_\text{Li}$

$\Delta m = [3(1.007825) + 4(1.008665)] - 7.000000$

$\Delta m = [3.023475 + 4.03466] - 7.000000$

$\Delta m = 7.058135 - 7.000000 = 0.058135\ \text{u}$

∴ Mass defect in the formation of $^7_3\text{Li}$ nucleus = 0.058135 u