Join thousands of students who trust us to help them ace their exams!
Multiple Choice
Provide a daughter nuclide when Rn-215 undergoes 2 sets of alpha decay and a positron emission.
A
B
C
D
0 Comments
Verified step by step guidance
1
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.