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226Ac can decay by any of three different nuclear processes: a emission, b emission, or electron capture. Write a balanced nuclear equation for the decay of 226Ac by each process.
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21. Nuclear Chemistry
Electron Capture & Positron Emission
Scelta multipla
Provide a daughter nuclide when Rn-215 undergoes 2 sets of alpha decay and a positron emission.
A
B
C
D
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Start by understanding the process of alpha decay. An alpha particle consists of 2 protons and 2 neutrons, which means it has a mass number of 4 and an atomic number of 2. When a nuclide undergoes alpha decay, it loses an alpha particle, reducing its mass number by 4 and its atomic number by 2.
Apply the first alpha decay to Rn-215. The mass number of Rn-215 is 215, and its atomic number is 86. Subtract 4 from the mass number and 2 from the atomic number to find the new nuclide after the first alpha decay.
Apply the second alpha decay to the resulting nuclide from the first decay. Again, subtract 4 from the mass number and 2 from the atomic number of the new nuclide to find the nuclide after the second alpha decay.
Understand positron emission. Positron emission occurs when a proton in the nucleus is converted into a neutron, releasing a positron. This decreases the atomic number by 1 but leaves the mass number unchanged.
Apply positron emission to the nuclide resulting from the second alpha decay. Decrease the atomic number by 1 while keeping the mass number the same to find the final daughter nuclide.
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