A nuclear fusion reaction is given by
`._(1)H^(2)+._(1)H^(2)rarr._(1)He^(3)+._(0)^(1)n + Q ("energy")`.
If `2` moles of deuterium are fused, then total released energy is

Explanation

(a) In a nuclear reaction, the sum of the masses of the target nucleus `(._(1)^(2)H)` and the bombarding particle `(._(1)^(2)H)` may be greater the product nucleus `(._(2)^(3)He)` and outgoing neutron `._(0)^(1)n`. So from the law of conservation of mass-energy some energy `(3.27 MeV)` is evolved due to mass defect in the nuclear reaction. This energy is called `Q`-value of the nuclear reaction.
(b) `"Density of the nuclues"=("mass of nucleus")/("Volume of nuclues")`
Mass of the nucleus`=A "amu"`
`=Axx1.66xx10^(-27)kg`
Volume of hte nucleus `=(4)/(3)=piR^(3)=(4)/(3)pi(R_(0)A^(1//3))^(3)=(4)/(3)piR_(0)^(3)A`
Thus, density `=(Axx1.66xx10^(-27))/(((4)/(3)piR_(0)^(3)))=(1.66xx10^(-27))/(((4)/(3)piR_(0)^(3)))`
Which shows that the density in independent of mass number `A`.
Using `R_(0)=1.1xx10^(-15)m`, we get
density `=2.297xx10^(17) kg m^(-3)`