Suppose a samarium-147 nuclide decays into a neodymium-143 nuclide by emitting an alpha particle.Complete the nuclear chemical equation below so that it describes this nuclear reaction.

The nuclear chemical equation for the decay of a samarium-147 nuclide into a neodymium-143 nuclide by emitting an alpha particle is:

147Sm --> 143Nd + 4He

Here's the step-by-step process:

1. Identify the atomic number and atomic mass of the parent nuclide (samarium-147). Samarium (Sm) has an atomic number of 62 and an atomic mass of 147.

2. Identify the atomic number and atomic mass of the daughter nuclide (neodymium-143). Neodymium (Nd) has an atomic number of 60 and an atomic mass of 143.

3. An alpha particle consists of 2 protons and 2 neutrons, so it has an atomic number of 2 and an atomic mass of 4.

4. In a nuclear reaction, the sum of the atomic numbers and the sum of the atomic masses on both sides of the equation must be equal (law of conservation of mass and charge).

5. Therefore, the atomic number on the left side of the equation (62 for Sm) must equal the sum of the atomic numbers on the right side of the equation (60 for Nd and 2 for the alpha particle). The atomic mass on the left side of the equation (147 for Sm) must equal the sum of the atomic masses on the right side of the equation (143 for Nd and 4 for the alpha particle).

6. So, the completed nuclear chemical equation is: 147Sm --> 143Nd + 4He.