Identify the unknown element 'X' in the following nuclear reactions: 1) $${ }^{10}_{5} B + X \rightarrow { }^{13}_{6} N + { }^{1}_{0} n$$ 2) $${ }^{9}_{4} B e + { }^{4}_{2} h e \rightarrow { }^{1}_{0} n + X$$ 3) $${ }^{27}_{13} A l + { }^{1}_{1} H \rightarrow { }^{23}_{11} N a + X$$

The provided image contains handwritten nuclear reactions. We will solve for the unknown element 'X' in each reaction by conserving the atomic and mass numbers. 1) $${ }^{10}_{5} B + X \rightarrow { }^{13}_{6} N + { }^{1}_{0} n$$ To find X, we balance the atomic numbers (subscripts) and mass numbers (superscripts) on both sides of the equation. Atomic number balance: $$5 + Z_X = 6 + 0 \Rightarrow Z_X = 1$$ Mass number balance: $$10 + A_X = 13 + 1 \Rightarrow A_X = 4$$ An element with atomic number 1 and mass number 4 is Deuterium ($${ }^{4}_{1} D$$), also known as Hydrogen-4, but this is not a standard stable isotope. If 'X' is meant to be a particle or a different element, let's re-examine the problem. There might be a typo in the image or the problem statement. Let's assume the first reaction is $${ }^{10}_{5} B + { }^{4}_{2} He \rightarrow { }^{13}_{6} N + { }^{1}_{0} n$$. In this case, $${ }^{4}_{2} He$$ (alpha particle) is one of the reactants. If X is not $${ }^{4}_{2} He$$, and the reaction is as written, then X would be $${ }^{4}_{1} D$$ (Deuterium) which is unusual. Assuming the problem meant to ask for a common reaction, let's check other possibilities for X. If X is an alpha particle ($${ }^{4}_{2} He$$), then $$5+2=7$$ and $$10+4=14$$, which does not match the products. If X is a neutron ($${ }^{1}_{0} n$$), then $$5+0=5$$ and $$10+1=11$$, which does not match. Let's consider the possibility that X is a proton ($${ }^{1}_{1} H$$). Then $$5+1=6$$ and $$10+1=11$$, which does not match the mass number of Nitrogen (13). Given the provided OCR and image, the most direct balancing gives $$Z_X = 1$$ and $$A_X = 4$$. This corresponds to Deuterium $${ }^{4}_{1}D$$. However, this is an uncommon representation in such problems. Let's proceed with the other reactions assuming standard conventions. If the intention was a standard reaction, the first reaction might be miswritten or X represents a less common particle. 2) $${ }^{9}_{4} B e + { }^{4}_{2} h e \rightarrow { }^{1}_{0} n + X$$ Atomic number balance: $$4 + 2 = 0 + Z_X \Rightarrow Z_X = 6$$ Mass number balance: $$9 + 4 = 1 + A_X \Rightarrow 13 = 1 + A_X \Rightarrow A_X = 12$$ An element with atomic number 6 is Carbon ($${ }_{6} C$$). Thus, $$X = { }^{12}_{6} C$$. 3) $${ }^{27}_{13} A l + { }^{1}_{1} H \rightarrow { }^{23}_{11} N a + X$$ Atomic number balance: $$13 + 1 = 11 + Z_X \Rightarrow 14 = 11 + Z_X \Rightarrow Z_X = 3$$ Mass number balance: $$27 + 1 = 23 + A_X \Rightarrow 28 = 23 + A_X \Rightarrow A_X = 5$$ An element with atomic number 3 is Lithium ($${ }_{3} L i$$). Thus, $$X = { }^{5}_{3} L i$$. Given the ambiguity in the first reaction, and assuming it's a standard nuclear physics problem, there might be a typo. If we strictly follow the balancing rules for the first reaction as written, X is $${ }^{4}_{1}D$$. However, if we consider common nuclear reactions involving Boron, it often interacts with alpha particles or neutrons. Without further clarification or correction of the first reaction, we provide the derived values. Final Answer Derivation: Reaction 1: $${ }^{10}_{5} B + X \rightarrow { }^{13}_{6} N + { }^{1}_{0} n$$. Balancing yields $$X = { }^{4}_{1}D$$. Reaction 2: $${ }^{9}_{4} B e + { }^{4}_{2} h e \rightarrow { }^{1}_{0} n + X$$. Balancing yields $$X = { }^{12}_{6} C$$. Reaction 3: $${ }^{27}_{13} A l + { }^{1}_{1} H \rightarrow { }^{23}_{11} N a + X$$. Balancing yields $$X = { }^{5}_{3} L i$$.