Balance the following eq equation: (}_{6)^14Carrow 0_(1)e+underline ( ) (}^14_{5)B (}_{2)^4He (}_{6)^14C {}^14N

To balance the nuclear reaction equation, we need to ensure that the sum of the atomic numbers (the subscript) and the sum of the mass numbers (the superscript) are equal on both sides of the equation.<br /><br />Given:<br />${}_{6}^{14}C\rightarrow 0_{1}e+\underline {\quad }$<br /><br />Let's analyze the options:<br /><br />1. ${}^{14}_{5}B$: This option has an atomic number of 5, which is incorrect since the atomic number of carbon is 6.<br />2. ${}_{2}^{4}He$: This option has an atomic number of 2, which is incorrect since the atomic number of carbon is 6.<br />3. ${}_{6}^{14}C$: This option is the same as the reactant, so it cannot be the product.<br />4. ${}^{14}N$: This option has an atomic number of 7, which is correct since the atomic number of carbon is 6.<br /><br />Therefore, the correct answer is:<br />${}^{14}N$<br /><br />The balanced nuclear reaction equation is:<br />${}_{6}^{14}C\rightarrow 0_{1}e+{}^{14}N$