Question

How much energy is released if 2.50 g of deuterons fuse with tritium and 50% of the neutrons are caputred by 6Li? (the 6 is supposed to be a superscript)

Answer 1

2 g of deutron = 1 mole of deutron = 6.022 x 10²³ Deutron

2.5 g of deutrons = 2.5/2 x 6.022 x 10²³ = 7.5275 x 10²³ deutrons

Energy produced by single DT reaction = 17.59 MeV

Hence, energy produced by 2.5 g of deutrons = 17.59 x 7.5275 x 10²³ MeV

E = 132.4 x 10²³ MeV

As each DT reaction produces a neutron with 14.1 MeV energy.

And half of the neutrons get captured in 6Li.

Hence, energy captured in 6Li = 1/2 x 7.5275 x 10²³ x 14.1 MeV

E_cap = 53.07 x 10²³ MeV

Hence, net energy released = (132.4 - 53.07) x 10²³ MeV

E_net = 79.33 x 10²³ MeV = 79.33 x 10²³ x 1.6 x 10^-13 J

E_net = 126.93 x 10¹⁰ J = 1269.3 GJ of energy.

This is by considering that the captured neutron in Li₆ do not react.

If they react, then we have to consider the energy produced by 1/2 x 7.5275 x 10²³ neutron Li₆ reactions also.

As each reaction may produce 4.78 Mev, the total energy may be 1/2 x 7.5275 x 10²³ x 4.78 MeV, in addition to 1269.3 GJ.