A proton with a kinetic energy `T- 1.5MeV` us captured by a deuteron `H^(2)`. Find the excitation energy of the formed nucleus.

Explanation

The reaction is `p+drarrHe^(3)`
Excitation energy of `He^(3)` is just the energy available in centre of mass. The velocity of the centre of mass is
`(sqrt(2m_(p)T_(p)))/(m_(p)+m_(d))~~(1)/(3)sqrt((2T_(p))/(m_(p)))`
In the `CM` frame, the kinetic enrgy available is `(m_(d)~~2m_(p))`
`(1)/(2)m_(p)((2)/(3)sqrt((2T_(p))/(m_(p))))^(2)+(1)/(2)2m_(p)((1)/(3)sqrt((2T)/(m_(p))))^(2)=(2T)/(3)`
The total energy available is then `Q=(2T)/(3)`
where `Q=c^(2)(Delta_(n)+Delta(d)-Delta_(He)^(3))`
`=c^(2)xx(0.00783+0.01410-0.01603)am u`
`= 5.49MeV`
Finally `E=6.49MeV`