The Q value of the Na23 (n,α) F20 reaction is -5.4MeV. Determine the threshold energy of the neutrons for this reaction (22.9898u, 18.9964u, 4.0039u, 1.008665u are the masses of Na23, F20 α and n respectively).
23Na+1n→20F+4He,^{23}Na+^1n\to^{20}F+^4He,23Na+1n→20F+4He,
Q=ΔE=(22.9898+1.008665−19.9964−4.0039)c2=−5.3 MeV.Q=\Delta E=(22.9898+1.008665-19.9964-4.0039)c^2=-5.3~MeV.Q=ΔE=(22.9898+1.008665−19.9964−4.0039)c2=−5.3 MeV.
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