Question

The mass of deuteron ${(}_1{H}^2)$ 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 ${}_1{H}^2$ nucleus is

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

Answer 1

Answer: (B)

Here, $m_p=1.007825u,m_n=1.008665u$
mass of ${}_1{H}^2$ nucleus, $m_N{(}_1{H}^2)=2.014102u$
The deuteron nucleus contains one proton and one neutron.
Therefore, mass of nucleons constituting deuteron,
$m_p+m_n=1.007825+1.008665=2.01649u$
Mass defect, $\Delta M=(m_p+m_n)-m_N{(}_1{H}^2)$
$=2.01649-2.014102=0.002388u$
$=0.002388\times 931.5MeV/c^2$
$=2.224MeV/c^2$Binding energy, $E_b=\Delta M{c}_2=2.224MeV$
Binding energy per nucleon
$=\frac{E_b}{A}=\frac{2.224}{2}=1.112MeV$