The equation is

_{3}Li^{7} + _{1}H^{1} —–→ _{4}Be^{7} + _{0}n^{1} + E

Now mass of _{3}Li^{7} = 7.01823 u

Mass of _{1}H^{1} = 1.00814 u

Mass on LHS = 7.01823 + 1.00814 = 8.02637 u

Mass of _{4}Be^{7} = 7.01592 u

Mass of _{0}n^{1} = 1.00866 u

Mass on RHS = 8.02458 u

Difference in masses, Δ m = 0.00179 u

When converted to energy, this will be the K.E of the proton.

0.00179 u = 931.5 × 0.00179 = **1.667 MeV.**

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