Question

given the nuclear reaction equation. $\frac{14}{7}n+\frac{4}{2}helongrightarrow\frac{a}{z}e+\frac{1}{1}h$. what are the values of a, z and what is the symbol of e. a= choose z= choose e= choose

Explanation:

Step1: Balance mass - number (A)

The sum of mass - numbers on the left - hand side of the nuclear reaction must equal the sum of mass - numbers on the right - hand side. The mass - number of $\frac{14}{7}N$ is 14 and of $\frac{4}{2}He$ is 4, and of $\frac{1}{1}H$ is 1. So, $14 + 4=A + 1$. Solving for $A$ gives $A=14 + 4-1=17$.

Step2: Balance atomic - number (Z)

The sum of atomic - numbers on the left - hand side of the nuclear reaction must equal the sum of atomic - numbers on the right - hand side. The atomic - number of $\frac{14}{7}N$ is 7 and of $\frac{4}{2}He$ is 2, and of $\frac{1}{1}H$ is 1. So, $7+2 = Z + 1$. Solving for $Z$ gives $Z=7 + 2-1=8$.

Step3: Identify the element

The element with atomic - number $Z = 8$ is oxygen, symbol $O$.

Answer:

$A = 17$
$Z = 8$
$E=O$