Answer :
Let's solve this step by step using the periodic table and the information provided.
### Step 1: Understand the Fusion Reaction for Helium
1. The first equation is:
[tex]\(\ce{_{1}^{2}H + _1^{3}H \rightarrow _{8}^{4}He}\)[/tex]
Here, the isotopes of hydrogen, [tex]\(\ce{^2H}\)[/tex] (deuterium) and [tex]\(\ce{^3H}\)[/tex] (tritium), are fusing to form an isotope of helium.
Let's analyze the masses and atomic numbers:
- The sum of the atomic numbers (protons) before the reaction is: [tex]\(1 + 1 = 2\)[/tex]
- The sum of the mass numbers before the reaction is: [tex]\(2 + 3 = 5\)[/tex]
In the product:
- The atomic number of Helium ([tex]\(\ce{He}\)[/tex]) is [tex]\(2\)[/tex]
- The mass number is [tex]\(4\)[/tex]
To balance the equation in terms of mass and atomic number, an additional neutron (mass number = 1, atomic number = 0) is needed:
[tex]\[\ce{_{1}^{2}H + _1^{3}H \rightarrow _{2}^{4}He + _{0}^1n}\][/tex]
### Step 2: Understand the Second Reaction Involving Nitrogen
2. The second equation is:
[tex]\[\ce{_{7}^{14}N + _{1}^1H \rightarrow _{Z}^{A}E}\][/tex]
Here, nitrogen ([tex]\(\ce{^7_{14}N}\)[/tex]) and hydrogen ([tex]\(\ce{^1_{1}H}\)[/tex]) react to form a new element [tex]\(E\)[/tex].
Let's analyze the masses and atomic numbers:
- The sum of the atomic numbers (protons) before the reaction is: [tex]\(7 + 1 = 8\)[/tex]
- The sum of the mass numbers before the reaction is: [tex]\(14 + 1 = 15\)[/tex]
Therefore, for the product [tex]\(E\)[/tex]:
- The atomic number [tex]\(Z\)[/tex] must be 8 (keeping the law of conservation of atomic numbers).
- The mass number [tex]\(A\)[/tex] must be 15 (keeping the law of conservation of mass numbers).
### Step 3: Identify the Element [tex]\(E\)[/tex]
Using the periodic table, we find:
- An element with atomic number 8 is oxygen ([tex]\(\ce{O}\)[/tex]).
Thus, the product [tex]\(E\)[/tex] is [tex]\(\ce{_8^{15}O}\)[/tex]:
The full reactions are:
1. [tex]\(\ce{_{1}^{2}H + _1^{3}H \rightarrow _{2}^{4}He + _{0}^1n}\)[/tex]
2. [tex]\(\ce{_{7}^{14}N + _{1}^1H \rightarrow _{8}^{15}O}\)[/tex]
The answers are therefore:
A: He
C: n
B: N
D: H
E: O
In written form:
- For the first fusion reaction: [tex]\( \ce{_1^2H + _1^3H \rightarrow _2^4He + _0^1n} \)[/tex]
- For the second reaction: [tex]\( \ce{_7^{14}N + _1^1H \rightarrow _8^{15}O } \)[/tex]
### Step 1: Understand the Fusion Reaction for Helium
1. The first equation is:
[tex]\(\ce{_{1}^{2}H + _1^{3}H \rightarrow _{8}^{4}He}\)[/tex]
Here, the isotopes of hydrogen, [tex]\(\ce{^2H}\)[/tex] (deuterium) and [tex]\(\ce{^3H}\)[/tex] (tritium), are fusing to form an isotope of helium.
Let's analyze the masses and atomic numbers:
- The sum of the atomic numbers (protons) before the reaction is: [tex]\(1 + 1 = 2\)[/tex]
- The sum of the mass numbers before the reaction is: [tex]\(2 + 3 = 5\)[/tex]
In the product:
- The atomic number of Helium ([tex]\(\ce{He}\)[/tex]) is [tex]\(2\)[/tex]
- The mass number is [tex]\(4\)[/tex]
To balance the equation in terms of mass and atomic number, an additional neutron (mass number = 1, atomic number = 0) is needed:
[tex]\[\ce{_{1}^{2}H + _1^{3}H \rightarrow _{2}^{4}He + _{0}^1n}\][/tex]
### Step 2: Understand the Second Reaction Involving Nitrogen
2. The second equation is:
[tex]\[\ce{_{7}^{14}N + _{1}^1H \rightarrow _{Z}^{A}E}\][/tex]
Here, nitrogen ([tex]\(\ce{^7_{14}N}\)[/tex]) and hydrogen ([tex]\(\ce{^1_{1}H}\)[/tex]) react to form a new element [tex]\(E\)[/tex].
Let's analyze the masses and atomic numbers:
- The sum of the atomic numbers (protons) before the reaction is: [tex]\(7 + 1 = 8\)[/tex]
- The sum of the mass numbers before the reaction is: [tex]\(14 + 1 = 15\)[/tex]
Therefore, for the product [tex]\(E\)[/tex]:
- The atomic number [tex]\(Z\)[/tex] must be 8 (keeping the law of conservation of atomic numbers).
- The mass number [tex]\(A\)[/tex] must be 15 (keeping the law of conservation of mass numbers).
### Step 3: Identify the Element [tex]\(E\)[/tex]
Using the periodic table, we find:
- An element with atomic number 8 is oxygen ([tex]\(\ce{O}\)[/tex]).
Thus, the product [tex]\(E\)[/tex] is [tex]\(\ce{_8^{15}O}\)[/tex]:
The full reactions are:
1. [tex]\(\ce{_{1}^{2}H + _1^{3}H \rightarrow _{2}^{4}He + _{0}^1n}\)[/tex]
2. [tex]\(\ce{_{7}^{14}N + _{1}^1H \rightarrow _{8}^{15}O}\)[/tex]
The answers are therefore:
A: He
C: n
B: N
D: H
E: O
In written form:
- For the first fusion reaction: [tex]\( \ce{_1^2H + _1^3H \rightarrow _2^4He + _0^1n} \)[/tex]
- For the second reaction: [tex]\( \ce{_7^{14}N + _1^1H \rightarrow _8^{15}O } \)[/tex]