27Mg can decay by beta minus emission to one of two possible excited states of 27Al - AQA - A-Level Physics - Question 31 - 2021 - Paper 2

To find the maximum possible kinetic energy of the beta particle emitted in this route, we need to consider the energy difference between the initial and final states during the beta decay process.

1. Identify the Energy Levels:

   • The energy associated with the ground state of 27Mg is given as 4.18 x 10^-13 J.
   • The two excited states have energies of 1.63 x 10^-13 J and 1.33 x 10^-13 J, respectively.

2. Calculate Energy Difference:

   • For the first route that emits a gamma photon corresponding to the higher energy state (1.63 x 10^-13 J):

     $E_{max} = E_{initial} - E_{final} = (4.18 imes 10^{-13} ext{ J}) - (1.63 imes 10^{-13} ext{ J})$

     $E_{max} = 2.55 imes 10^{-13} ext{ J}$

3. Kinetic Energy of the Beta Particle:

   • This energy difference corresponds to the kinetic energy of the emitted beta particle, which is the maximum possible value given this route, as energy conservation indicates that the total energy before the decay is equal to the sum of all energies after the decay.

4. Conclusion:

   • Therefore, the maximum possible kinetic energy for the beta particle emitted in this route is:

     Answer option C: 2.55 x 10^-13 J.