Find the energy of the reaction $N^{14}(\mathrm{\alpha }, \mathrm{p})O^{17}$ if the kinetic energy of the incoming alpha-particle is $T_{\mathrm{\alpha }}=4.0 \mathrm{MeV}$ and the proton is outgoing at an angle $\mathrm{\theta }=60°$ to the motion. The direction of the alpha-particle has a kinetic energy, $T_{\mathrm{p}}=2.09 \mathrm{MeV}$.

Excess of the mass of neutron proton and $\alpha$ particle are$∆{\mathrm{m}}_{\mathrm{p}}=0.00783 ∆{\mathrm{m}}_{\mathrm{\alpha }}=0.00260 \mathrm{amu} \mathrm{and} ∆{\mathrm{m}}_{\mathrm{n}}=0.00867 \mathrm{amu}$

$1 \mathrm{amu}=1.67\times {10}^{-24} \mathrm{g}$ and $1 \mathrm{amu}$ is equivalent to $931.5 \mathrm{MeV}$ of energy.

Excess of the mass of atoms $∆{\mathrm{Li}}^7=0.01601 \mathrm{amu}, ∆{\mathrm{Be}}^{10}=0.01294 \mathrm{amu}$ and of neutron and $\alpha -\mathrm{particle}$ are$∆{\mathrm{m}}_{\alpha }=0.00260 \mathrm{amu} \mathrm{and} ∆{\mathrm{m}}_{\mathrm{n}}=0.00867 \mathrm{amu}$

$1 \mathrm{amu}=1.67\times {10}^{-24} \mathrm{g}$ and $1 \mathrm{amu}$ is equivalent to $931.5 \mathrm{MeV}$ of energy.

Excess of the mass of atoms $∆{\mathrm{Li}}^7=0.01601 \mathrm{amu}, ∆{\mathrm{Be}}^7=0.01693 \mathrm{amu}$ and of neutron and proton are$∆{\mathrm{m}}_{\mathrm{p}}=0.00783 \mathrm{amu} \mathrm{and} ∆{\mathrm{m}}_{\mathrm{n}}=0.00867 \mathrm{amu}$

$1 \mathrm{amu}=1.67\times {10}^{-24} \mathrm{g}$ and $1 \mathrm{amu}$ is equivalent to $931.5 \mathrm{MeV}$ of energy.

Excess of the mass of atoms $∆{\mathrm{Li}}^7=0.01601 \mathrm{amu}, ∆{\mathrm{He}}^4=0.00260 \mathrm{amu}$ and $∆\mathrm{H}=0.00783 \mathrm{amu}$ and of neutron and proton are$∆{\mathrm{m}}_{\mathrm{p}}=0.00783 \mathrm{amu}$ and $∆{\mathrm{m}}_{\mathrm{n}}=0.00867 \mathrm{amu}$.

$1 \mathrm{amu}=1.67\times {10}^{-24} \mathrm{g}$ and $1 \mathrm{amu}$ is equivalent to $931.5 \mathrm{MeV}$ of energy.