Answer
a. The decay of 232Th (Z=90) to 236U (Z=92) + $\alpha$ is not possible because it would violate the conservation of both the atomic number (Z) and the mass number (A).
b. The decay of 238Pu (Z=94) to 236U (Z=92) + alpha is also not possible.
c. The decay of 11B (Z=5) to 11B (Z=5) + gamma is not possible. This decay is called isomeric transition or isomeric shift.
d. The decay of 33P (Z=15) to 32S (Z=16) + e- is not possible.
Work Step by Step
a. In this decay, the atomic number of the parent nucleus (232Th) is 90, while the atomic number of the daughter nucleus (236U) is 92. The alpha particle is also composed of two protons and two neutrons, so it also has an atomic number of 2. Adding up the atomic numbers of the daughter nucleus and the alpha particle would give 92 + 2 = 94, which is greater than the atomic number of the parent nucleus (90). This means that the decay would violate the conservation of the atomic number, and is therefore not possible.
b. The sum of mass number of daughter nucleus and alpha particle 236+4 = 240 which is greater than parent's mass number(238), So it is not possible.
c. There is no conservation of mass and atomic numbers, therefore it's not possible
d. This beta decay generally occurs when a nucleus has an excess of neutrons in relation to its protons and is trying to reach stability by emitting a beta particle (an electron or a positron) . The emission of a beta particle increases the atomic number of the nucleus by 1. So it is not possible for 33P to emit a beta particle and become 32S as it would increase its atomic number.
