Question

Calculate the energy released in the nuclear reaction \[\ce{_3^7Li + p ->2\alpha}\] given mass of \[\ce{_3^7Li}\] atom and of helium atom to be 7.016 u and 4.0026 u respectively.

Answer 1

Data: M₁ \[\ce{(_3^7Li atom)}\] = 7.016 u, M₂ = (He atom)

= 4.0026 u, m_p = 1.00728 u, 1 u = 931.5 MeV/c²

Δ M = M₁ + m_p - 2M₂

= (7.016 + 1.00728- 2(4.0026)]u

= 0.01808 u = (0.01808)(931.5) MeV/c²

= 16.84152 MeV/c² 

Therefore, the energy released in the nuclear reaction= (ΔM)c² = 16.84152 MeV