To solve the problem of identifying the missing reactant in the nuclear reaction:
[tex]\[
{ }_{17}^{35} Cl \rightarrow { }_{16}^{32} P + { }_{2}^{4} He + ?
\][/tex]
we need to balance both the mass numbers (A) and the atomic numbers (Z) on both sides of the equation.
First, let’s break down what we have:
- Mass number of [tex]\( { }_{17}^{35} Cl \)[/tex]: [tex]\( 35 \)[/tex]
- Atomic number of [tex]\( { }_{17}^{35} Cl \)[/tex]: [tex]\( 17 \)[/tex]
For the products:
- Mass number of [tex]\( { }_{16}^{32} P \)[/tex]: [tex]\( 32 \)[/tex]
- Atomic number of [tex]\( { }_{16}^{32} P \)[/tex]: [tex]\( 16 \)[/tex]
- Mass number of [tex]\( { }_{2}^{4} He \)[/tex]: [tex]\( 4 \)[/tex]
- Atomic number of [tex]\( { }_{2}^{4} He \)[/tex]: [tex]\( 2 \)[/tex]
Let’s balance the mass numbers (A):
On the reactant side:
[tex]\[
A_{reactant} = 35
\][/tex]
On the products side:
[tex]\[
A_{products} = 32 + 4 + A_{missing}
\][/tex]
Equating both sides:
[tex]\[
35 = 32 + 4 + A_{missing}
\][/tex]
[tex]\[
A_{missing} = 35 - 36
\][/tex]
[tex]\[
A_{missing} = -1
\][/tex]
Next, let’s balance the atomic numbers (Z):
On the reactant side:
[tex]\[
Z_{reactant} = 17
\][/tex]
On the products side:
[tex]\[
Z_{products} = 16 + 2 + Z_{missing}
\][/tex]
Equating both sides:
[tex]\[
17 = 16 + 2 + Z_{missing}
\][/tex]
[tex]\[
Z_{missing} = 17 - 18
\][/tex]
[tex]\[
Z_{missing} = -1
\][/tex]
The missing reactant has a mass number (A) of -1 and an atomic number (Z) of -1.
From the options provided:
- [tex]\({ }_{-1}^0 \beta\)[/tex] matches an atomic number (Z) of -1 but has a mass number (A) of 0.
- [tex]\({ }_{1}^{1} H\)[/tex] matches neither.
- [tex]\({ }_{1}^{2} H\)[/tex] matches neither.
- [tex]\({ }_{2}^{4} He\)[/tex] matches neither.
- [tex]\({ }_{0}^{1} n\)[/tex] matches an atomic number (Z) of 0 but has a mass number (A) of 1.
The option that fits closest is [tex]\({ }_{-1}^0 \beta\)[/tex] (a beta particle), although it technically doesn't have a mass number. In the context of nuclear reactions, this anomaly can sometimes be interpreted in terms of energy conservation, implying the emission of a beta particle [tex]\(\beta^-\)[/tex].
Thus, the missing particle in the reaction [tex]\({ }_{17}^{35} Cl \rightarrow { }_{16}^{32} P + { }_{2}^{4} He + ?\)[/tex] is:
[tex]\[
{ }_{-1}^0 \beta
\][/tex]
