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(A) Given: Mass of neutron $m_n = 1.00893 \ amu$,mass of proton $m_p = 1.00813 \ amu$,and mass of deuteron $m_d = 2.01473 \ amu$. $A$ deuteron nucleus

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$11.65 	imes 10^{-4}$

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(A) Given: Mass of neutron $m_n = 1.00893 \ amu$,mass of proton $m_p = 1.00813 \ amu$,and mass of deuteron $m_d = 2.01473 \ amu$. $A$ deuteron nucleus $(^2_1H)$ consists of one proton and one neutron. The mass defect $\Delta m$ is calculated as: $\Delta m = (m_n + m_p) - m_d$ $\Delta m = (1.00893 + 1.00813) - 2.01473$ $\Delta m = 2.01706 - 2.01473 = 0.00233 \ amu$. The packing fraction is defined as the ratio of mass defect to the mass number $(A)$. For deuteron,$A = 2$. $	ext{Packing fraction} = \frac{\Delta m}{A} = \frac{0.00233}{2} = 0.001165 = 11.65 	imes 10^{-4}$.

The masses of neutron,proton and deuteron in amu are $1.00893$,$1.00813$ and $2.01473$,respectively. The packing fraction of the deuteron in amu is

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