Let's analyze the nuclear equation step-by-step to determine which particle completes the nuclear equation:
Given the initial nucleus:
[tex]\[
{}_{56}^{124} Ba
\][/tex]
This nucleus decays into:
[tex]\[
{}_{55}^{124} Cs
\][/tex]
To determine the emitted particle, we need to consider the conservation of the atomic number (Z) and the mass number (A):
1. Mass Number (A):
The mass number before the decay is 124. The mass number of the daughter nucleus (Cesium) is also 124. Therefore, the mass number of the emitted particle must be 0 to maintain the conservation of mass.
2. Atomic Number (Z):
The atomic number before the decay is 56 (for Barium). The atomic number of the daughter nucleus (Cesium) is 55. This means the atomic number has decreased by 1.
So, we have:
[tex]\[
56 (Ba) \rightarrow 55 (Cs) + Z_{\text{emitted\_particle}}
\][/tex]
To balance the atomic number, the emitted particle must carry an atomic number of:
[tex]\[
56 - 55 = 1
\][/tex]
However, the actual change we need is a decrease of 1 in the atomic number, which implies the emitted particle should have an atomic number of -1 to balance the decrease:
[tex]\[
56 - (-1) = 55
\][/tex]
Combining these two observations (mass number of 0 and atomic number of -1), the emitted particle is:
[tex]\[
{}_{-1}^0 e
\][/tex]
This is the notation for a beta particle (an electron), which has a mass number of 0 and an atomic number of -1.
Thus, the particle that completes the nuclear equation is:
[tex]\[
{}_{-1}^0 e
\][/tex]
