Complete the table on the basis of information available in the symbols given below:
$(a)$ ${ }_{17}^{35} Cl$
$(b)$ ${ }_{6}^{12} C$
$(c)$ ${ }_{35}^{81} Br$

Element	$n_{(p)}$	$n_{(n)}$

(N/A) To find the number of protons $(n_{(p)})$ and neutrons $(n_{(n)})$, we use the notation ${ }_{Z}^{A} X$, where $Z$ is the atomic number (number of protons) and $A$ is the mass number (protons + neutrons).
$1$. For ${ }_{17}^{35} Cl$: $n_{(p)} = Z = 17$. $n_{(n)} = A - Z = 35 - 17 = 18$.
$2$. For ${ }_{6}^{12} C$: $n_{(p)} = Z = 6$. $n_{(n)} = A - Z = 12 - 6 = 6$.
$3$. For ${ }_{35}^{81} Br$: $n_{(p)} = Z = 35$. $n_{(n)} = A - Z = 81 - 35 = 46$.

Element	$n_{(p)}$	$n_{(n)}$
$Cl$	$17$	$18$
$C$	$6$	$6$
$Br$	$35$	$46$