Answer :
Let's fill in the missing numbers and symbols for the given nuclear processes step by step:
### Part A
The original nuclear equation is:
[tex]\[ \ce{_{16}^{31} Si \longrightarrow{ }_{14}^{28} Si +{ }_0^1 n } \][/tex]
To balance the equation, we need to find the missing isotope that decays into [tex]\(\ce{_{14}^{28} Si}\)[/tex] and [tex]\(\ce{_0^1 n}\)[/tex].
The isotope [tex]\(\ce{_{16}^{31} Si}\)[/tex] has:
- Mass number = [tex]\(31\)[/tex]
- Atomic number = [tex]\(16\)[/tex]
The given product [tex]\(\ce{_{14}^{28} Si}\)[/tex] has:
- Mass number = [tex]\(28\)[/tex]
- Atomic number = [tex]\(14\)[/tex]
A neutron [tex]\(\ce{_0^1 n}\)[/tex] has:
- Mass number = [tex]\(1\)[/tex]
- Atomic number = [tex]\(0\)[/tex]
We apply the conservation of mass number and atomic number:
- Mass number: [tex]\(31 = 28 + 1\)[/tex]
- Atomic number: [tex]\(16 = 14 + 0\)[/tex]
So the original equation is balanced correctly.
### Part B
The original nuclear equation is:
[tex]\[ \ce{_{19}^{40} K \longrightarrow{ }_{18}^{40} Ar +{ }_1^1 H } \][/tex]
For potassium [tex]\(\ce{_{19}^{40} K}\)[/tex]:
- Mass number = [tex]\(40\)[/tex]
- Atomic number = [tex]\(19\)[/tex]
For argon [tex]\(\ce{_{18}^{40} Ar}\)[/tex]:
- Mass number = [tex]\(40\)[/tex]
- Atomic number = [tex]\(18\)[/tex]
For the proton [tex]\(\ce{_1^1 H}\)[/tex]:
- Mass number = [tex]\(1\)[/tex]
- Atomic number = [tex]\(1\)[/tex]
We check for the conservation of mass number and atomic number:
- Mass number: [tex]\(40 = 40 + 1\)[/tex]
- Atomic number: [tex]\(19 = 18 + 1\)[/tex]
So the original equation is balanced correctly.
### Part C
The original nuclear equation is:
[tex]\[ \ce{_{24}^{52} Cr \longrightarrow{ }_{22}^{48} Ti +[]^{[]}[] } \][/tex]
For chromium [tex]\(\ce{_{24}^{52} Cr}\)[/tex]:
- Mass number = [tex]\(52\)[/tex]
- Atomic number = [tex]\(24\)[/tex]
For titanium [tex]\(\ce{_{22}^{48} Ti}\)[/tex]:
- Mass number = [tex]\(48\)[/tex]
- Atomic number = [tex]\(22\)[/tex]
We need to find the missing particle that completes the equation. Let's denote it by [tex]\(\ce{_z^A X}\)[/tex]:
Applying the conservation laws:
- Mass number: [tex]\(52 = 48 + A\)[/tex]
- Atomic number: [tex]\(24 = 22 + z\)[/tex]
Solving for [tex]\(A\)[/tex] (mass number) and [tex]\(z\)[/tex] (atomic number):
- [tex]\(A = 52 - 48 = 4\)[/tex]
- [tex]\(z = 24 - 22 = 2\)[/tex]
This missing particle is [tex]\(\ce{_2^4 He}\)[/tex] (an alpha particle).
So the completed equation is:
[tex]\[ \ce{_{24}^{52} Cr \longrightarrow { }_{22}^{48} Ti +_{2}^{4} He } \][/tex]
### Part D
The original nuclear equation is:
[tex]\[ \ce{_{24}^{55} Cr \longrightarrow{ }_{25}^{55} Mn +{ }_{-1}^0 \beta } \][/tex]
For chromium [tex]\(\ce{_{24}^{55} Cr}\)[/tex]:
- Mass number = [tex]\(55\)[/tex]
- Atomic number = [tex]\(24\)[/tex]
For manganese [tex]\(\ce{_{25}^{55} Mn}\)[/tex]:
- Mass number = [tex]\(55\)[/tex]
- Atomic number = [tex]\(25\)[/tex]
For the beta particle [tex]\(\ce{_{-1}^0 \beta}\)[/tex]:
- Mass number = [tex]\(0\)[/tex]
- Atomic number = [tex]\(-1\)[/tex]
We check for the conservation of mass number and atomic number:
- Mass number: [tex]\(55 = 55 + 0\)[/tex]
- Atomic number: [tex]\(24 = 25 - 1\)[/tex]
So the original equation is balanced correctly.
Combining all the parts, we have:
- Part A: [tex]\[ \ce{_{16}^{31} Si \longrightarrow{ }_{14}^{28} Si +{ }_0^1 n } \][/tex]
- Part B: [tex]\[ \ce{_{19}^{40} K \longrightarrow{ }_{18}^{40} Ar +{ }_1^1 H } \][/tex]
- Part C: [tex]\[ \ce{_{24}^{52} Cr \longrightarrow{ }_{22}^{48} Ti +_{2}^{4} He } \][/tex]
- Part D: [tex]\[ \ce{_{24}^{55} Cr \longrightarrow{ }_{25}^{55} Mn +{ }_{-1}^0 \beta } \][/tex]
### Part A
The original nuclear equation is:
[tex]\[ \ce{_{16}^{31} Si \longrightarrow{ }_{14}^{28} Si +{ }_0^1 n } \][/tex]
To balance the equation, we need to find the missing isotope that decays into [tex]\(\ce{_{14}^{28} Si}\)[/tex] and [tex]\(\ce{_0^1 n}\)[/tex].
The isotope [tex]\(\ce{_{16}^{31} Si}\)[/tex] has:
- Mass number = [tex]\(31\)[/tex]
- Atomic number = [tex]\(16\)[/tex]
The given product [tex]\(\ce{_{14}^{28} Si}\)[/tex] has:
- Mass number = [tex]\(28\)[/tex]
- Atomic number = [tex]\(14\)[/tex]
A neutron [tex]\(\ce{_0^1 n}\)[/tex] has:
- Mass number = [tex]\(1\)[/tex]
- Atomic number = [tex]\(0\)[/tex]
We apply the conservation of mass number and atomic number:
- Mass number: [tex]\(31 = 28 + 1\)[/tex]
- Atomic number: [tex]\(16 = 14 + 0\)[/tex]
So the original equation is balanced correctly.
### Part B
The original nuclear equation is:
[tex]\[ \ce{_{19}^{40} K \longrightarrow{ }_{18}^{40} Ar +{ }_1^1 H } \][/tex]
For potassium [tex]\(\ce{_{19}^{40} K}\)[/tex]:
- Mass number = [tex]\(40\)[/tex]
- Atomic number = [tex]\(19\)[/tex]
For argon [tex]\(\ce{_{18}^{40} Ar}\)[/tex]:
- Mass number = [tex]\(40\)[/tex]
- Atomic number = [tex]\(18\)[/tex]
For the proton [tex]\(\ce{_1^1 H}\)[/tex]:
- Mass number = [tex]\(1\)[/tex]
- Atomic number = [tex]\(1\)[/tex]
We check for the conservation of mass number and atomic number:
- Mass number: [tex]\(40 = 40 + 1\)[/tex]
- Atomic number: [tex]\(19 = 18 + 1\)[/tex]
So the original equation is balanced correctly.
### Part C
The original nuclear equation is:
[tex]\[ \ce{_{24}^{52} Cr \longrightarrow{ }_{22}^{48} Ti +[]^{[]}[] } \][/tex]
For chromium [tex]\(\ce{_{24}^{52} Cr}\)[/tex]:
- Mass number = [tex]\(52\)[/tex]
- Atomic number = [tex]\(24\)[/tex]
For titanium [tex]\(\ce{_{22}^{48} Ti}\)[/tex]:
- Mass number = [tex]\(48\)[/tex]
- Atomic number = [tex]\(22\)[/tex]
We need to find the missing particle that completes the equation. Let's denote it by [tex]\(\ce{_z^A X}\)[/tex]:
Applying the conservation laws:
- Mass number: [tex]\(52 = 48 + A\)[/tex]
- Atomic number: [tex]\(24 = 22 + z\)[/tex]
Solving for [tex]\(A\)[/tex] (mass number) and [tex]\(z\)[/tex] (atomic number):
- [tex]\(A = 52 - 48 = 4\)[/tex]
- [tex]\(z = 24 - 22 = 2\)[/tex]
This missing particle is [tex]\(\ce{_2^4 He}\)[/tex] (an alpha particle).
So the completed equation is:
[tex]\[ \ce{_{24}^{52} Cr \longrightarrow { }_{22}^{48} Ti +_{2}^{4} He } \][/tex]
### Part D
The original nuclear equation is:
[tex]\[ \ce{_{24}^{55} Cr \longrightarrow{ }_{25}^{55} Mn +{ }_{-1}^0 \beta } \][/tex]
For chromium [tex]\(\ce{_{24}^{55} Cr}\)[/tex]:
- Mass number = [tex]\(55\)[/tex]
- Atomic number = [tex]\(24\)[/tex]
For manganese [tex]\(\ce{_{25}^{55} Mn}\)[/tex]:
- Mass number = [tex]\(55\)[/tex]
- Atomic number = [tex]\(25\)[/tex]
For the beta particle [tex]\(\ce{_{-1}^0 \beta}\)[/tex]:
- Mass number = [tex]\(0\)[/tex]
- Atomic number = [tex]\(-1\)[/tex]
We check for the conservation of mass number and atomic number:
- Mass number: [tex]\(55 = 55 + 0\)[/tex]
- Atomic number: [tex]\(24 = 25 - 1\)[/tex]
So the original equation is balanced correctly.
Combining all the parts, we have:
- Part A: [tex]\[ \ce{_{16}^{31} Si \longrightarrow{ }_{14}^{28} Si +{ }_0^1 n } \][/tex]
- Part B: [tex]\[ \ce{_{19}^{40} K \longrightarrow{ }_{18}^{40} Ar +{ }_1^1 H } \][/tex]
- Part C: [tex]\[ \ce{_{24}^{52} Cr \longrightarrow{ }_{22}^{48} Ti +_{2}^{4} He } \][/tex]
- Part D: [tex]\[ \ce{_{24}^{55} Cr \longrightarrow{ }_{25}^{55} Mn +{ }_{-1}^0 \beta } \][/tex]
