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  • The mass of deuteron  nucleus is <img alt="2.014102 \, \, \mathrm{u}" s

The mass of deuteron \left({ }_1 \mathrm{H}^2\right) nucleus is 2.014102 \, \, \mathrm{u}. If the masses of proton and neutron are 1.007825\, \, \mathrm{u} and 1.008665\, \, \mathrm{u} respectively, then the binding energy per nucleon of { }_1 \mathrm{H}^2 nucleus is:

Option: 1

2.2 MeV


Option: 2

1.1 MeV


Option: 3

0.5 MeV


Option: 4

0.25 MeV


Answers (1)

best_answer

Here,  \mathrm{m}_{\mathrm{p}}=1.007825 \mathrm{u}, \mathrm{m}_{\mathrm{n}}=1.008665 \mathrm{u}

\text { mass of }{ }_1 \mathrm{H}^2 \text { nucleus, } \mathrm{m}_{\mathrm{N}}\left({ }_1 \mathrm{H}^2\right)=2.014102 \mathrm{u}

The deuteron nucleus contains one proton and one neutron.

Therefore, mass of nucleons constituting deuteron,

\mathrm{m}_{\mathrm{p}}+\mathrm{m}_{\mathrm{n}}=1.007825+1.008665=2.01649 \, \mathrm{u}

\text { Mass defect, } \Delta M=\left(m_p+m_n\right)-m_N\left({ }_1 H^2\right)

=2.01649-2.014102=0.002388 \mathrm{u}

=0.002388 \times 931.5 \, \mathrm{MeV} / \mathrm{c}^2

=2.224 \, \mathrm{MeV} / \mathrm{c}^2

Binding energy, \mathrm{E}_{\mathrm{b}}=\Delta \mathrm{Mc}^2=2.224\, \mathrm{MeV}

Binding energy per nucleon =\frac{\mathrm{E}_{\mathrm{b}}}{\mathrm{A}}=\frac{2.224}{2}=1.112 \, \mathrm{MeV}

Posted by

manish painkra

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