Question

For each of the isotopes in Exercise 21.1, determine the numbers of protons, neutrons, and electrons in a neutral atom of the isotope.

(a) ${}_{11}^{24}Na$

(b) ${}_{13}^{29}Al$

(c) $73Kr$

(d) ${}_{77}^{194}Ir$

Short answer

a) The number of protons and electrons is 11, and the number of neutrons is $13.$

b) The number of protons and electrons is 13, and the number of neutrons is $16.$

c) The number of protons and electrons is 36, and the number of neutrons is $37.$

d) The number of protons and electrons is 77, and the number of neutrons is $117.$

Step-by-step solution

Introduction

The general representation of an element is:$\frac{\mathbf{A}}{\mathbf{Z}}\mathbf{X}$, where Z is the atomic number of element and A is the mass number.

Atomic number of element is the number of protons and it is equal to the number of electrons in an atom.

Mass number is the total number of protons and neutron in an atom.

Isotopes are atoms of the same element which have the same atomic number (Z), but different mass number (A). This means they have different number of neutrons.

$\begin{array}{r}\mathbf{Z}\mathbf{=}\mathbf{N}\left({\mathbf{p}}^{\mathbf{+}}\right)\mathbf{=}\mathbf{N}\left({\mathbf{e}}^{\mathbf{-}}\right)\\ \mathbf{A}\mathbf{=}\mathbf{N}\left({\mathbf{p}}^{\mathbf{+}}\right)\mathbf{+}\mathbf{N}\left({\mathbf{n}}^{\mathbf{0}}\right)\end{array}$

Hence,

$\mathbf{N}\left({\mathbf{n}}^{\mathbf{0}}\right)\mathbf{=}\mathbf{A}\mathbf{-}\mathbf{N}\left({\mathbf{p}}^{\mathbf{+}}\right)$

We have${}_{11}^{24}Na$.

$\begin{array}{rl}Z& =N\left(p^{+}\right)\\ & =N\left(e^{-}\right)\\ & =11\end{array}$

$\begin{array}{rl}N\left(n^0\right)& =A-N\left(p^{+}\right)\\ & =24-11\\ & =13\end{array}$

The number of protons and electrons is 11, and the number of neutrons is 13.

Subpart b)

The given nuclide is ${}_{13}^{29}Al$.

$\begin{array}{rl}Z& =N\left(p^{+}\right)\\ & =N\left(e^{-}\right)\\ & =13\\ N\left(n^0\right)& =A-N\left(p^{+}\right)\\ & =29-13\\ & =16\end{array}$

The number of protons and electrons is 13, and the number of neutrons is $16.$

Subpart c)

We have ${}_{36}^{73}Kr$

$\begin{array}{rl}Z& =N\left(p^{+}\right)\\ & =N\left(e^{-}\right)\\ & =36\\ N\left(n^0\right)& =A-N\left(p^{+}\right)\\ & =73-36\\ & =37\end{array}$

The number of protons and electrons is 36, and the number of neutrons is 37.

Subpart d)

We have${}_{77}^{194}\mathrm{I}\mathrm{r}$.

$\begin{array}{rl}Z& =N\left(p^{+}\right)\\ & =N\left(e^{-}\right)\\ & =77\\ N\left(n^0\right)& =A-N\left(p^{+}\right)\\ & =194-77\\ & =117\end{array}$