nuclear physics- Energetics and Mechanics of Nuclear Reaction Homework 
A sample of $^{24}_{12}\mathrm{Mg}$ is bombarded by a monoenergetic proton. If the resulting nucleus in a  $^{24}_{12}\mathrm{Mg}(p,\gamma)$ reaction; $^{24}_{12}\mathrm{Mg}(p,\gamma)$ has its first energy level at 2.5 MeV. What is the minimum energy of the of the proton to excite this level?


As I understand this problem, I assume that this is non-relativistic. I transformed the reaction from:
$^{24}_{12}\mathrm{Mg}(p,\gamma)$
into:
$^{24}_{12}\mathrm{Mg} + ^{1}_{1}\mathrm{H} \to \gamma + ^{25}_{13}\mathrm{Al}$
I found out that the product is not existing (because there is no 25,13 aluminum in our given table)
The formula for non-relativistic would be:
$$Q= KE_b + KE_y - KE_a + E$$
as part of my understanding to this problem, the 2.5 MeV will be the $KE_y$ and the variable to be find will be the internal energy ($E$).
Solving for $Q$
Q= (summation of excess mass of resultant) - (summation of excess mass of product)
Q= (-13.933 + 7.289) - (0)

i made the summation of excess mass of product into 0 because gamma has no excess mass(in our table) and the product is not existing (in our table)
Q= -6.644 MeV


KEb = Q / (1+ [mass of b / mass of Y]) 
    = -6.644 / (1 + 0)
    = -6.644 MeV

from the formula of non-relativistic:
E = -6.644 + 6.644 - 2.5
E = -2.5 MeV

for the velocity, as my professor did on her example, 
momentum (p)= mV
myVy = mbVb
Vy = (mbVb) / my

but in this problem, the unknown is Va (velocity of a or velocity of proton)
I am confused in this problem because of the terms "monoenergetic proton" and "first energy level" which is not that familiar to us. I am doubtful to how I understand the problem. Does my understanding correct? does my internal energy (E) correct? i don't know how to find the velocity. please help me to solve this problem.. this will be passed the same day we are going to take exam on this topic. our homework serve as our reviewer so please, I don't want to be confused in our examination day. 
 A: Just because $^{25}$Al isn't in your table doesn't mean it doesn't exist or hasn't been measured. Do a search for "mass of Al-25" or "chart of nuclides".  The mass (in u) will be there. 1u = 931.5 Mev/c$^2$. What table are you using that doesn't have $^{25}$Al?
Rather than dealing with excess masses I have found it easier to use
$Q = \epsilon_{reactants} - \epsilon_{products}$  where $\epsilon$ is the sum of the mass energies (mc$^2$), excited states or photons. Q is the reaction energy. If Q>0, then you get energy out. If Q<0, you have to put energy in. $Q$ is also equal to $K_{final}-K_{initial}$ where $K$ is the kinetic energy. $Q$ is separate from the energy required to overcome the Coulomb barrier (the energy required to bring the proton to contact the nucleus).
Another aspect you should investigate is conserving the invariant center-of-mass energy, $E^2-p^2c^2$. That quantity must be the same for the before ($^{24}$Mg + $^1$H) and after ($^{25}$Al + $\gamma$) systems. Here $E$ is the sum of the mass energy and kinetic energy of a particle.
