4) When Neptunium decays plutonium 239 is formed. a. What kind of particle is emitted? B Write the equation.

## Step 1
The problem involves the concept of nuclear decay, specifically alpha decay. In alpha decay, an atomic nucleus emits an alpha particle, which is a helium nucleus (2 protons and 2 neutrons). This process reduces the atomic number by 2 and the mass number by 4.

## Step 2
In this case, Neptunium (Np) with an atomic number of 93 and a mass number of 237 is decaying to form Plutonium (Pu) with an atomic number of 94 and a mass number of 239. The difference in atomic numbers and mass numbers between the parent and daughter nuclei is due to the emission of an alpha particle.

## Step 3
The nuclear decay equation can be written as follows:

### $Np^{237}_{93} \rightarrow Pu^{239}_{94} + \alpha^{4}_{2}$

Here, $Np^{237}_{93}$ represents the parent nucleus (Neptunium), $Pu^{239}_{94}$ represents the daughter nucleus (Plutonium), and $\alpha^{4}_{2}$ represents the emitted alpha particle.