Problem 10: Find out the K.E to which a proton must be accelerated to induce the following nuclear reaction. Li7 (p, n) Be7.
The equation is
3Li7 + 1H1 —–→ 4Be7 + 0n1 + E
Now mass of 3Li7 = 7.01823 u
Mass of 1H1 = 1.00814 u
Mass on LHS = 7.01823 + 1.00814 = 8.02637 u
Mass of 4Be7 = 7.01592 u
Mass of 0n1 = 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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