A neutron causes the element 88X232 to change according to the reaction 0n1 + 88X232 ---->ZYA + . Identify the (a) atomic number and (b) atomic mass number of the daughter nucleus. (c) This product subsequently undergoes - decay, and its daughter does too. Identify the atomic number of the final nucleus. (d) Give the atomic mass number of this final nucleus.
Any help here would be much appriciated! I don't even understand the equations
0n1 + 88X232 ---->ZYA
element 88 is radium Ra
0n1 + 88Ra232 ---->88Ra233
There is either infromation missing in the question or is it Ra223 you are starting with?
If so the question makes more sense
0n1 + 88Ra223 ---->88Ra224
and Ra 224 under goes alpha decay to
88Ra224 -> 2He4 + 86Rn220
(half life is about 3.7 days and is part of the well known thorium 232 decay series).
Sure, let's break down the reaction and answer each part of your question step by step.
The given nuclear reaction is: 0n1 + 88X232 → ZYA + .
(a) To identify the atomic number of the daughter nucleus (ZYA), we need to understand that the atomic number represents the number of protons in an atom's nucleus. In this reaction, the neutron (0n1) is absorbed by the starting nuclide (88X232), resulting in the formation of a new nuclide (ZYA). Since neutrons have no charge, the formation of the daughter nucleus does not involve any protons. Therefore, the atomic number of the daughter nucleus (ZYA) remains the same as the atomic number of the starting nuclide (88X232), which is 88.
(b) To identify the atomic mass number of the daughter nucleus (ZYA), we need to understand that the atomic mass number represents the total number of protons and neutrons in an atom's nucleus. Since the neutron (0n1) is absorbed by the starting nuclide (88X232), the total number of protons and neutrons in the daughter nucleus (ZYA) would be equal to the atomic mass number of the starting nuclide (88X232), which is 232.
(c) The given question states that the daughter nucleus subsequently undergoes β- decay (or electron emission). In this type of decay, a neutron within the nucleus is converted into a proton while emitting an electron and an antineutrino. Since a neutron changes into a proton during β- decay, the atomic number of the nucleus increases by one. Therefore, the atomic number of the final nucleus would be 89.
(d) Although the question does not explicitly mention the atomic mass number of the final nucleus, we can infer it from the information provided. We know that the parent nuclide (88X232) has an atomic mass number of 232. During β- decay, the atomic number increases by one, but the atomic mass number remains the same. Therefore, the atomic mass number of the final nucleus would also be 232.
To summarize:
(a) The atomic number of the daughter nucleus (ZYA) is 88.
(b) The atomic mass number of the daughter nucleus (ZYA) is 232.
(c) The atomic number of the final nucleus is 89.
(d) The atomic mass number of the final nucleus is 232.