To solve for [tex]\( M \)[/tex] in each of the nuclear reaction equations, we need to understand that the subscripts represent the atomic number (number of protons) and the superscripts represent the mass number (total number of protons and neutrons) of each nucleus or particle. We can balance the equations by ensuring both the atomic numbers and mass numbers are the same on both sides of each equation.
Let's solve them one by one:
(i) [tex]\({ }_7^{14} N + \frac{1}{2} \mu c \rightarrow \frac{17}{17} + M\)[/tex]
We balance both sides:
- Mass number: [tex]\(14 + 0.5 = 17 + M_{\text{mass}}\)[/tex]
[tex]\(M_{\text{mass}} = 14.5 - 17 = -2.5\)[/tex]
Since mass number must be positive, we must reconsider the context or the provided numbers as traditional nuclear reactions do not normally involve fractional atomic or mass numbers.
(ii) [tex]\(II + 3L \rightarrow 2M\)[/tex]
This equation is written in an unconventional format and lacks clear atomic and mass numbers. With insufficient information, determining [tex]\(M\)[/tex] isn't possible.
(iii) [tex]\({ }_4 H B_c + { }_2^4 H_e \rightarrow M\)[/tex]
Let's assume [tex]\(B\)[/tex] has an atomic number of 4 and a mass number of [tex]\(A_1\)[/tex], and He has an atomic number of 2 and a mass number of 4.
- Atomic number: [tex]\(4 + 2 = M_{\text{atomic}}\)[/tex]
[tex]\(M_{\text{atomic}} = 6\)[/tex]
- Mass number: [tex]\(A_1 + 4 = M_{\text{mass}}\)[/tex]
[tex]\(M_{\text{mass}} = A_1 + 4\)[/tex]
Without specifics on [tex]\(A_1\)[/tex], the equation to match [tex]\(A_1\)[/tex] can’t be balanced fully.
(iv) [tex]\({ }_{15}^{30} P \rightarrow M + -9 e\)[/tex]
An electron ([tex]\(e^-\)[/tex]) has a charge -1 and negligible mass.
- Atomic number: [tex]\(15 = M_{\text{atomic}} + (-9)\)[/tex]
[tex]\(M_{\text{atomic}} = 15 + 9 = 24\)[/tex]
- Mass number: [tex]\(30 = M_{\text{mass}}\)[/tex]
[tex]\(M_{\text{mass}} = 30\)[/tex]
Based on this reaction, [tex]\(M\)[/tex] would be [tex]\({}_{24}^{30}\)[/tex].
(v) [tex]\({ }_7^{35} Cl + { }_0^1 n \rightarrow M + { }_1^1 H\)[/tex]
Balancing the numbers:
- Atomic number: [tex]\(7 + 0 = M_{\text{atomic}} + 1\)[/tex]
[tex]\(M_{\text{atomic}} = 7 - 1 = 6\)[/tex]
- Mass number: [tex]\(35 + 1 = M_{\text{mass}} + 1\)[/tex]
[tex]\(M_{\text{mass}} = 35\)[/tex]
Thus, from this reaction, [tex]\(M\)[/tex] is [tex]\({}_{6}^{35}\)[/tex].
Summarizing:
(i) Interpretation of the non-integer and hypothetical nature.
(ii) Insufficient information to determine [tex]\(M\)[/tex].
(iii) Without a specific mass number (Assumed notation, more specific [tex]\(B_c\)[/tex]).
(iv) [tex]\({}_{24}^{30} (possibly)\)[/tex]
(v) \({}_{6}^{35} (possible element, unknown notation)
