Nuclear Reaction - How to determine variables as much as possible

[Chris89]

Hey Guys,
exercise: "It is desired to study the first excited state of 16O which is at energy of 6.049 MeV.
Using the (alpha, n) reaction on target of 13C, what is the minimum energy of incident alphas which will populate the excited state?

So, i suggest to define first the reaction equation:
alpha + 13C -> 16O + n where E(16O) = 6,049 MeV

Because we are searching the minimum energy of the alpha, i would define the Q value = 0
(with kinetic energy of particle i = Ti)

-> Tc + Ta = To +Tn
-> Tc + Ta - Tn = 6,049 MeV

How i get more information about the system?
I could use a change of the system from lab-frame(momentum of 13C = 0) to Center of Mass frame(Sum(momentum) = 0)), but i can't see how we determine there more variables with our given informations.

Does somebody know a "trick" to get there more informations of the system to finally get the min. E of alphas?

Best Regards,
Chris

 

[gleem]

To review. The Q-Value is the difference in the total mass -energy of the initial state and that of the final state when the final nucleus state is in its ground state(GS). For this reaction it is +2.216 MeV. If Q value is positive it is the excess mass-energy in the final state. Thus if the projectile particle with no kinetic energy is joined with the target leaving the resultant nuclei in the ground state then 2.2156 Mev is available to be share as kinetic energy between the recoiling O¹⁶ and the emitted neutron. If it is negative it is the kinetic energy which the projectile must have which will join the two leaving in the resulting nucleus (in GS) and outgoing particle with no kinetic energy.

Using the notate m and M for ground state masses, p for projectile, t for target, r for resulting nucleus , e for emitted particle, T for kinetic energy and E for energy of an excited state we define the Q-value as

Q = (m_p + M_t)c² - ( m_e + M_r)c²

Conservation of energy give the equation

(m_p + M_t)c² + T_p = (M_r + m_e)c²+ E_r + T_r + T_e

Thus we can write

Q = T_r + E_r + T_e - T_p

For the minimum projectile energy to populate the excited state solve for Tp and set Te and Tr equal to zero. (Why?)

Thus Tp = Ex - Q. Thus the minimum is projectile energy is 6.049 - 2.216 = 3.833 MeV