# Answer to Question #21450 in Atomic and Nuclear Physics for Bekezela Moyo

Question #21450
protons of energy 3 MeV bombard Li7 and neutrons are emitted from the target at right angles to the incident beam of protons. The Q-value of the reaction 3Li7 (p,n)4Be7 is -1.8 MeV. (1) Calculate the energy of the emitted neutrons. (2) Derive the expression of the Q-value you use.
In nuclear physics and chemistry, the Q value for areaction is the amount of energy released by that reaction:

Q = E(reactants) - E(products)
A reaction with a positive Q value is exothermic (has anet release of energy), while a reaction with a negative Q value is endothermic
(requires a net energy input) In our case:
Q = 3 MeV + (M_p + M_li)*c^2 - (M_n + M_be)*c^2 - E_kn -E_kbe
where, E_kn - kinetic energy of the neutron E_kbe -kinetic energy of the Be-7.
We can also write this equation though the Bindingenergy.Q = 3 MeV - E_b(Li) + E_b(Be) - E_kn - E_kbe = 3 Mev -32.9 Mev + 37.6 Mev - E_kn - E_kbe = 1.8 Mev =&gt; =&gt; E_kn + E_kbe = 5.9 MeV
We also need to use conversation of the momentum.
vee P_p = vec P_n+ vec P_Be; vec - vector; we alsoremember, that P =sqrt(E *2m) (non relativistic case, because energies are too
low )
along x axis:
P_p = 3 MeV * 2 * u = E_kbe*2*u* cos^2(alpha) along y:
E_kbe*2*u*sin^2(alpha) = E_kn*2*u
So, we have three equations and three unknowns.
E_kn + E_kbe = 5.9 MeV =&gt; E_kn + E_kn/(7*sin^2(alpha))= 5.9 MeV =&gt; E_kn
+ E_kn/(7*sin(acrtan(E_kn/3))) = 5.9 MeV
E_kbe*7*sin^2(alpha) = E_kn
3 MeV = E_kbe*7*cos^2(alpha) =&gt; tan^2(alpha) = E_kn/3
E_kn + E_kn/(7*sin(acrtan(sqrt(E_kn/3)))) = 5.9 MeV -From this equation wee will find E_kn.
E_kn = 4.99 MeV (Maple 16 was used here)

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