Question

Calculate the $Q$-value of the reaction;
${}_3^{6}Li{+}_0^{1}n{⟶}_2^{4}He{+}_1^{3}H$
Given:

${}_3^{6}Li=6.015126amu{,}_2^{4}He=4.002604amu$

${}_1^{3}H=3.016049amu{,}_0^{1}n=1.008665amu$

Answer 1

$Q=(6.015126+1.008665-3.016049-4.002604)\times 931.5Mev$

$Q=5.138\times {10}^{-13}\times 931.5Mev$

$=4.786Mev$