Question

Consider the D-T reaction (deuterium-tritium fusion)
$$_{ 1 }^{ 2 }{ Ce }+_{ 1 }^{ 3 }{ Ru }\rightarrow _{ 2 }^{ 4 }{ He }+n$$
Calculate the energy released in $$MeV$$ in this reaction from the data:
$$m\left( _{ 1 }^{ 2 }{ H } \right) =2.014102u;\quad m\left( _{ 1 }^{ 3 }{ H } \right) =3.016049u$$

Answer 1

The D-T reaction is

$$_{ 1 }^{ 2 }{ H }+_{ 1 }^{ 3 }{ H }\rightarrow _{ 2 }^{ 4 }{ He }+_{ 0 }^{ 1 }{ n }+Q$$

If $${ m }_{ N }\left( _{ 1 }^{ 2 }{ H } \right) ,{ m }_{ N }\left( _{ 1 }^{ 3 }{ H } \right) ,{ m }_{ N }\left( _{ 2 }^{ 4 }{ He } \right) $$ represent nuclear masses

$$Q=\left[ { m }_{ N }\left( _{ 1 }^{ 2 }{ H } \right) +{ m }_{ N }\left( _{ 1 }^{ 3 }{ H } \right) -{ m }_{ N }\left( _{ 2 }^{ 4 }{ He } \right) -{ m }_{ N } \right] \times 931.5MeV...(1)$$

Here instead of nuclear masses, the masses of $$_{ 1 }^{ 2 }{ H },_{ 1 }^{ 3 }{ H },_{ 2 }^{ 4 }{ He }$$ atoms have been given. Therefore, nuclear masses of $$_{ 1 }^{ 2 }{ H },_{ 1 }^{ 3 }{ H },_{ 2 }^{ 4 }{ He }$$ atoms may be found by subtracting mass of $$1$$ electron and $$2$$ electrons respectively from their atomic masses

If $${ m }_{ }\left( _{ 1 }^{ 2 }{ H } \right) ,{ m }_{ }\left( _{ 1 }^{ 3 }{ H } \right) ,{ m }_{ }\left( _{ 2 }^{ 4 }{ He } \right) $$ represent atomic masses, then equation (1) becomes

$$Q=\left[ \left\{ m\left( _{ 1 }^{ 2 }{ H } \right) -{ m }_{ e } \right\} +\left\{ m\left( _{ 1 }^{ 3 }{ H } \right) -{ m }_{ e } \right\} -\left\{ m\left( _{ 2 }^{ 4 }{ He } \right) -2{ m }_{ e } \right\} -{ m }_{ n } \right] \times 931.5MeV$$

$$=\left[ 2.014102+3.016049-4.002603-1.008665 \right] \times 931.5MeV=17.59MeV$$