Question

The mass of deuteron $(_1H^2)$ nucleus is 2.014102 u. If the masses of proton and neutron are 1.007825 u and 1.008665 u respectively, then the binding energy per nucleon of $_1H^2$ nucleus is

• A. 2.2 MeV
• B. 1.1 MeV
• C. 0.5 MeV
• D. 0.25 MeV

Answer 1

Answer: (B)

Here, $m_p = 1.007825\, u, \,m_n = 1.008665\, u$
mass of $_1H^2$ nucleus, $m_N(_1H^2) = 2.014102\, u$
The deuteron nucleus contains one proton and one neutron.
Therefore, mass of nucleons constituting deuteron,
$m_p + m_n = 1.007825 + 1.008665 = 2.01649\, u$
Mass defect, $ΔM = (m_p + m_n) - m_N(_1H^2)$
$= 2.01649 - 2.014102 = 0.002388\, u$
$= 0.002388 × 931.5\, MeV/c^2$
$= 2.224 \,MeV/c^2$Binding energy, $E_b = ΔMc_2 = 2.224\, MeV$
Binding energy per nucleon
$= \frac{E_{b}}{A} = \frac{2.224}{2} = 1.112\,MeV$