Mg can decay by beta minus emission to one of two possible excited states of Al - AQA - A-Level Physics - Question 31 - 2021 - Paper 2

To find the maximum possible kinetic energy of the beta particle emitted, we need to consider the difference in energy between the excited state and the ground state.

1. Identify the energies involved:
   The excited states are given as 1.33 x 10^{-13} J and 1.63 x 10^{-13} J. The ground state energy (final energy) is 4.18 x 10^{-13} J.

2. Calculate the energy released:
   The maximum kinetic energy of the beta particle emitted will be the difference between the energy of the excited state and the ground state energy of Al. Hence:

   • For the excited state of 1.33 x 10^{-13} J:

     $KE_{max} = E_{excited} - E_{ground} = 1.33 imes 10^{-13} J - 0.00 = 1.33 imes 10^{-13} J$

   • For the excited state of 1.63 x 10^{-13} J:

     $KE_{max} = 1.63 imes 10^{-13} J - 0.00 = 1.63 imes 10^{-13} J$

3. Determine the maximum kinetic energy:
   The maximum possible kinetic energy for the beta particle emitted corresponds to the higher excited state energy of 1.63 x 10^{-13} J since it directly influences the kinetic energy calculation.

Thus, the answer will be given as:

Answer: 1.63 x 10^{-13} J.