Write balanced nuclear equations for the following:
`beta`-emission from magnesium-`28`

Answer

To write the balanced nuclear equation for the beta emission from magnesium-28, we follow these steps:

### Step 1: Identify the initial isotope
The initial isotope is magnesium-28, which has:
- Mass number (A) = 28
- Atomic number (Z) = 12

### Step 2: Write the nuclear symbol for magnesium-28
The nuclear symbol for magnesium-28 can be written as:
\[ \text{ }^{28}_{12}\text{Mg} \]

### Step 3: Understand beta emission
Beta emission involves the conversion of a neutron into a proton, which results in the emission of a beta particle (an electron). The beta particle is represented as:
\[ \text{ }^{0}_{-1}\text{e} \]

### Step 4: Determine the new atomic number and mass number
During beta emission:
- The mass number remains unchanged (A = 28).
- The atomic number increases by 1 (Z = 12 + 1 = 13).

### Step 5: Identify the product isotope
The new element formed after beta emission will have an atomic number of 13, which corresponds to aluminum (Al). Thus, the nuclear symbol for aluminum-28 is:
\[ \text{ }^{28}_{13}\text{Al} \]

### Step 6: Write the balanced nuclear equation
Now, we can combine all the components to write the balanced nuclear equation for the beta emission from magnesium-28:
\[ \text{ }^{28}_{12}\text{Mg} \rightarrow \text{ }^{28}_{13}\text{Al} + \text{ }^{0}_{-1}\text{e} \]

### Final Balanced Equation
The final balanced nuclear equation is:
\[ \text{ }^{28}_{12}\text{Mg} \rightarrow \text{ }^{28}_{13}\text{Al} + \text{ }^{0}_{-1}\text{e} \]

---