The yeild of the nuclear reaction `C^(13)(d,n)N^(14)` has maximum magnitudes at the following values of kinetic energy `T_(i)` of bombarding deuterons: `0,60,0.90, 1.55,`and `1.80MeV`. Making use of the tabel of atomic masses, find the corresponding energy levels of the transitional nucleus through which this reaction proceeds.

Explanation

The reaction is
`d+C^(13rarrN^(15))rarrn+N^(14)`
Maxima of yields determine the energy levels of `N^(15)`. As in the previous problem the excitation enrgy is
`E_(exc)=Q+E_(K)`
where `E_(k)=` available kinetic enrgy. This is found is as in the previous problem. The velocity of the centre of mass is
`(sqrt(2m_(d)T_(i)))/(m_(d)+m_(c ))=(m_(d))/(m_(d)+m_(c ))sqrt((2Ti)/(m_(d)))`
So `E_(k)=(1)/(2)m_(d)(1-(m_(d))/(m_(d)+m_(c ))^(2)(2Ti)/(m_(d)))+(1)/(2)m_(c )((m_(d))/(m_(d)+m_(c )))^(2)(2Ti)/(m_(d))=(m_(c ))/(m_(d)+m_(c ))Ti`
`Q` is the `Q` value for the ground state of `N^(15)`: we have
`Q=c^(2)xx(Delta_(d)+Delta_(c )^(13)-Delta_(N)^(15))`
`1=c^(2)xx(0.01410+0.00335-0.00011)am u`
`=16.14MeV`
The excitaiton energies then are
`16.66MeV,16.92MeV`
`17.49MeV` and `17.70MeV`.