What is the amount of energy released by deuterium and tritum fusion ?

Options

A. 60.6 eV
B. 12.6 eV
C. 17.6 eV
D. 28.3 eV

Answer

C

Explanation

`underset("Deuterium")(._(1)^(2)H)+underset("Tntium")(._(1)^(3)H)to underset("Helium")(._(2)^(4)He)+underset("neutron")(._(0)^(1)n)+Q`
The energy released in the process is given by
`Q=[M_(._(1)^(2)H)+M_(._(1)^(3)H)-M_(._(2)^(4)He)-M_(._(0)^(1)n)]c^(2)`
`=[2.014102+3.016050-4.002603-1.008665]uc^(2)`
`=(0.018884 u)[931.5(MeV)/(u)]=17.6 MeV`.