The answer is will require 2.2 MeV of energy
The minimum energy required to knock o the neutron out of a deuteron atom is equal to the binding energy of the atom
[tex]$B \cdot E=\Delta m c^{2} \quad[\Delta m=$[/tex] mass defect [tex]$]$[/tex]
[tex]\Delta m=\left(m_{\text {Proton }}+m_{\text {Nutron }}\right)-m_{\text {atom }}.[/tex]
[tex]$\begin{aligned} \Delta m &=m_{P}+m_{N}-m_{a t o m} \\ &=(1.007276+1.008665-2.013553) \mathrm{u} \\ &=0.002388 \mathrm{u} \\ \Delta m &=0.002388 \times 1.661 \times 10^{-27} 0 . \mathrm{kg} \\ &=0.003966 \times 10^{-27 \mathrm{lgg}} \\ &=3.966 \times 10^{-30} \mathrm{~kg} \end{aligned}$[/tex]
[tex]$\begin{aligned} B \cdot E &=\Delta m c^{2} \\ &=3.966 \times 10^{-30} \times\left(3 \times 10^{8}\right)^{2} \\ &=35.694 \times 10^{-14} \mathrm{~J} \\ &=3.5694 \times 10^{-13} \mathrm{~J} . \end{aligned}$[/tex]
[tex]$B \cdot E=\frac{3.5694 \times 10^{-13}}{1.6 \times 10^{-19} \times 10^{-6}} \mathrm{MeV}$[/tex]
[tex]B \cdot E=2.23 \mathrm{MeV}[/tex]
What is binding energy?
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