An alpha particle with kinetic energy \(7.70 \mathrm{MeV}\) strikes a \({ }^{14} \mathrm{~N}\) nucleus at rest. An \({ }^{17} \mathrm{O}\) nucleus and a proton are produced, the proton emitted at \(90^{\circ}\) to the direction of the incident alpha particle and carrying kinetic energy \(4.44 \mathrm{MeV}\). The rest energies of the various particles are: alpha particle, \(3730.4 \mathrm{MeV}\); \({ }^{14} \mathrm{~N}, 13,051 \mathrm{MeV} ;\) proton, \(939.29 \mathrm{MeV} ;{ }^{17} \mathrm{O}, 15,843 \mathrm{MeV}\) (a) Find the kinetic energy of the \({ }^{17} \mathrm{O}\) nucleus. (b) At what angle with respect to the direction of the incident alpha particle does the \({ }^{17} \mathrm{O}\) nucleus move?

Step by step solution

01

Determine Initial and Final Energies and Momenta

The initial kinetic energy is the kinetic energy of the alpha particle which is 7.70 MeV. The rest energies of all the particles before and after the collision are given. Sum up the rest energies and kinetic energies before and after the collision. Note that before the collision, only the alpha particle has kinetic energy. After the collision, both the oxygen nucleus and the proton have kinetic energy.

02

Calculate the Kinetic Energy of the Oxygen Nucleus

Conservation of energy gives us \[E_{initial}=E_{final}\], where \(E_{initial}\) is the total initial energy and \(E_{final}\) is the total final energy. By subtracting the energies after the collision not related to the kinetic energy of the oxygen nucleus from both sides, the kinetic energy of the oxygen nucleus can be determined: \[E_{oxygen,KE}=E_{final}-E_{initial}+7.7 MeV - 4.44 MeV\]. Add the kinetic energy of the alpha particle from before the collision and subtract the kinetic energy of the proton after the collision.

03

Determine the Direction of the Oxygen Nucleus

For the direction of the oxygen nucleus, we can use the conservation of momentum. The momentum of the alpha particle before the collision was in a certain direction. After the collision, the momenta of the proton and oxygen nucleus together must have the same magnitude and direction. Given that the proton moves at 90 degrees to the original direction, the direction of the oxygen nucleus can be computed by considering right triangle geometry with the sides of the triangle being the momenta of the proton and oxygen nucleus and the alpha particle momentum being the hypotenuse.

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