Numerical

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.

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#### Solution

**Data: **M_{1} \[\ce{(_3^7Li atom)}\] = 7.016 u, M_{2} = (He atom)

= 4.0026 u, m_{p} = 1.00728 u, 1 u = 931.5 MeV/c^{2}

Δ M = M_{1} + m_{p} - 2M_{2}

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

= 0.01808 u = (0.01808)(931.5) MeV/c^{2}

= 16.84152 MeV/c^{2}^{ }

Therefore, the energy released in the nuclear reaction= (ΔM)c^{2} = 16.84152 MeV

Concept: Nuclear Energy

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