When an alpha particle ⁴He) with a kinetic energy of 6.54 MeV strikes a ²⁷Al nucleus, a ³⁰P nucleus is created and a neutron is ejected. Assume that the ³⁰P nucleus has no final kinetic energy. Given the following masses for the various particles, what is the kinetic energy of the ejected neutron?

Question

Quizplus · Accepted Answer

Using conservation of mass and energy: Initial mass: ⁴He + ²⁷Al Final mass: ³⁰P + neutron Initial mass = (4.0026 u + 126.9045 u) c^2 Final mass = (129.9177 u + m_n) c^2 Conservation of mass: 4.0026 u + 126.9045 u = 129.9177 u + m_n m_n = 1.9894 u Conservation of energy: Initial energy – final energy = kinetic energy of ejected neutron (6.54 MeV/nucleus x 2 nuclei) – 0 = KE_neutron KE_neutron = 6.54 MeV Therefore, the kinetic energy of the ejected neutron is 6.54 MeV, which is choice A.

When an Alpha Particle 4He) with a Kinetic Energy of 6.54

When an Alpha Particle ⁴He) with a Kinetic Energy of 6.54