Certainly! Let's consider the nuclear equation:
[tex]\[ {}_{11}^{22} \text{Na} \longrightarrow {}_{10}^{22} \text{Ne} + {}_a^0 \beta \][/tex]
In this nuclear equation, we have sodium-22 decaying into neon-22 and a beta particle. The challenge is to find the missing value "a" that balances the equation.
### Concepts Used:
1. Conservation of Mass Number (A): The total mass number before and after the decay must be the same.
2. Conservation of Atomic Number (Z): The total atomic number before and after the decay must be the same.
### Given Values:
- Initial nucleus (sodium-22):
- Mass number (A): 22
- Atomic number (Z): 11
- Resultant nucleus (neon-22):
- Mass number (A): 22
- Atomic number (Z): 10
- Beta particle:
- Mass number (A): 0 (beta particles almost do not affect the mass number)
### Step-by-Step Solution:
1. Balance the Mass Number (A):
- Sodium-22 has a mass number of 22.
- Neon-22 has a mass number of 22.
- The beta particle has a mass number of 0.
- Thus, the mass numbers are already balanced:
[tex]\[
22 \longrightarrow 22 + 0
\][/tex]
2. Balance the Atomic Number (Z):
- Sodium-22 has an atomic number of 11.
- Neon-22 has an atomic number of 10.
- The beta particle must compensate for the difference in atomic numbers.
[tex]\[
11 \longrightarrow 10 + a
\][/tex]
3. Solve for the Missing Value (a):
- Initially, we have 11 protons from sodium.
- After decay, we need to end up with 11 total protons, which includes protons from both neon and the beta particle.
- Hence:
[tex]\[
11 \longrightarrow 10 + a \implies a = 1
\][/tex]
The missing value [tex]\(a\)[/tex] that balances the equation is the charge of a [tex]\(\beta\)[/tex]-minus (β⁻) particle, which has a charge of [tex]\(-1\)[/tex].
So, the missing value that will balance the equation is [tex]\(-1\)[/tex].
Thus, the correct option is:
[tex]\[ \boxed{-1} \][/tex]
