To solve these nuclear fusion equations, we'll use the law of conservation of mass and atomic numbers. This law states that the sum of the mass numbers (A) and the sum of the atomic numbers (Z) on both sides of the equation must be equal.
### First Equation:
[tex]\[\mathrm{_1^2 H + \; _1^3 H \rightarrow \; _Z^A He}\][/tex]
1. Identify the mass numbers and atomic numbers for the reactants:
- [tex]\(\mathrm{_1^2 H}\)[/tex]: Mass number [tex]\(= 2\)[/tex], Atomic number [tex]\(= 1\)[/tex]
- [tex]\(\mathrm{_1^3 H}\)[/tex]: Mass number [tex]\(= 3\)[/tex], Atomic number [tex]\(= 1\)[/tex]
2. Add the mass numbers and atomic numbers of the reactants to get the mass and atomic numbers of the product:
- Total mass number [tex]\(= 2 + 3 = 5\)[/tex]
- Total atomic number [tex]\(= 1 + 1 = 2\)[/tex]
3. For the product [tex]\(\mathrm{He}\)[/tex] (Helium), the atomic number must be 2 because Helium's atomic number is 2 (indicated by the periodic table).
4. Since Helium's atomic number should be 2, the mass number [tex]\(A\)[/tex] of Helium must equal the total mass number (5) obtained from the sum of reactants.
Thus, the first equation is:
[tex]\[\mathrm{_1^2 H + \; _1^3 H \rightarrow \; _2^5 He}\][/tex]
Correspondingly:
- [tex]\( \boxed{2} \)[/tex] for Atomic number (Z) of Helium
### Second Equation:
[tex]\[\mathrm{_7^{14} N + \; _1^1 H \rightarrow \; _Z^A E}\][/tex]
1. Identify the mass numbers and atomic numbers for the reactants:
- [tex]\(\mathrm{_7^{14} N}\)[/tex]: Mass number [tex]\(= 14\)[/tex], Atomic number [tex]\(= 7\)[/tex]
- [tex]\(\mathrm{_1^1 H}\)[/tex]: Mass number [tex]\(= 1\)[/tex], Atomic number [tex]\(= 1\)[/tex]
2. Add the mass numbers and atomic numbers of the reactants:
- Total mass number [tex]\(= 14 + 1 = 15\)[/tex]
- Total atomic number [tex]\(= 7 + 1 = 8\)[/tex]
3. Since the addition of two elements on the reactant side forms a new compound plus a leftover particle,
- For the product [tex]\(E\)[/tex], the choices are typically [tex]\(\mathrm{_1^1 H}\)[/tex], [tex]\(\mathrm{_0^1 n}\)[/tex], or [tex]\(\mathrm{_1^2 H}\)[/tex].
Since the total mass number and atomic numbers must match:
[tex]\[\mathrm{_7^{14} N + \; _1^1 H \rightarrow \; _8^{15} O}\][/tex]
[tex]\[\mathrm{_0 E}\][/tex] must be a particle that, when added to [tex]\(\mathrm{_8^{15} O}\)[/tex], balances the above reaction.
Here E is a beta particle typically denoted as [tex]\(\mathrm{_1^0 e}\)[/tex] or [tex]\(_{+1}^0 e^+ \)[/tex] [positron], or [tex]\(_{-1}^0 e^- \)[/tex] [electron].
Since:
[tex]\[ \mathrm{_7^{14} N + \; _1^1 H \rightarrow \; _8^{15} O + \; _{+1}^0 e^+} \][/tex]
Thus,
- [tex]\(A = 15\)[/tex]
- [tex]\(B = 8\)[/tex]
- [tex]\(C = 15\)[/tex]
- [tex]\(D = 8\)[/tex]
- [tex]\(E = e^+\)[/tex] (positron)
Therefore:
[tex]\[
\begin{aligned}
\boxed{2} \quad &\text{(Atomic number of Helium in first equation)} \\
\boxed{8} \quad &\text{(Atomic number of Oxygen in second equation)}
\end{aligned}
\][/tex]
