Determine the maximum kinetic energy of the beta particle emitted during the following decay scheme: ¹² N ¹² C ^ * + e⁺ + ¹² C ^ * ¹² C + ;(4.43 MeV ) . The atomic mass of ¹² N is given as 12.018613 u .

Step-by-step solution

The maximum kinetic energy of the emitted positron ( e⁺ ) is equal to the Q -value of the decay process. For a ⁺ decay, the Q -value in terms of atomic masses is given by: Q = [m(¹² N ) - m(¹² C ^ * ) - 2m_e]c^2 Since the ¹² C ^ * nucleus is in an excited state and subsequently emits a gamma ray of energy E_ = 4.43 MeV to reach the ground state, its mass can be written as: m(¹² C ^ * )c^2 = m(¹² C )c^2 + E_ Substituting this into the Q -value equation, we get: K_ = Q = [m(¹² N ) - m(¹² C )]c^2 - 2m_e c^2 - E_ Given