Answer to Question #21450 in Atomic and Nuclear Physics for Bekezela Moyo
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 => => 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 => E_kn + E_kn/(7*sin^2(alpha))= 5.9 MeV => 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) => 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)