Question

Find number of protons and neutrons for (a),(b) and (c) respectively.

(a) $\begin{array}{c}35\\ 17\end{array}Cl$(b) $\begin{array}{c}12\\ 6\end{array}C$(c) $\begin{array}{c}81\\ 35\end{array}Br$.

• A. $n_p$= 17,6,35
  $n_n$= 18,6,46
• B. $n_p$=16,6,35
  $n_n$= 18,6,46
• C. $n_p$=17,6,35
  $n_n$= 18,6,45
• D. $n_p$=16,6,35
  $n_n$= 18,6,45

Answer 1

The correct option is A $n_p$= 17,6,35

$n_n$= 18,6,46
For an element $X$, the mass number and atomic no. are represented as; $X_{atomicno}^{massno}$

Number of neutrons + number of protons= Atomic mass number.

Atomic number = number of protons = number of electrons
Therefore for $Cl$: $n_p$ $=17$ & $n_n$ $=35-17=18$
For $C$ : $n_p$ $=6$ & $n_n$ $=12-6=6$
For $Br$: $n_p$ $=35$ & $n_n$ $=81-35=46$