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27 Mg can decay by beta minus emission to one of two possible excited states of 27 Al - AQA - A-Level Physics - Question 31 - 2021 - Paper 2
Question 31
27 Mg can decay by beta minus emission to one of two possible excited states of 27 Al. Both excited states decay by the emission of a gamma photon directly to the g... show full transcript
Worked Solution & Example Answer:27 Mg can decay by beta minus emission to one of two possible excited states of 27 Al - AQA - A-Level Physics - Question 31 - 2021 - Paper 2
Step 1
What is the maximum possible kinetic energy for the beta particle emitted in this route?
Answer
To find the maximum possible kinetic energy of the beta particle emitted during the beta decay process, we can use the energy difference between the initial and final states.
The excited state of 27 Mg has energy levels of 1.63 x 10^{-13} J and 1.33 x 10^{-13} J. The ground state for 27 Al has an energy of 0.00 J. The path that involves the highest energy difference indicates the potential maximum kinetic energy output.
Calculating the maximum kinetic energy:
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The energy difference for the route emitting a gamma photon with the higher frequency:
E_{max} = E_{initial} - E_{final} = (4.18 imes 10^{-13} ext{ J}) - (1.33 imes 10^{-13} ext{ J})
E_{max} = 2.85 imes 10^{-13} ext{ J}
Thus, the maximum possible kinetic energy for the beta particle emitted in this route is:
D. 2.85 x 10^{-13} J