Tabulate the number of neutrons, protons, and electrons for neutral atoms of $$\text { the following nuclides: } { } ^ { 29 } \mathrm { Si } , { } ^ { 44 } \mathrm { Sc } , { } ^ { 49 } \mathrm {~V} , { } ^ { 210 } \mathrm { Bi } \text {, and } { } ^ { 241 } \mathrm { Pu } \text {. }$$

Question

Quizplus · Accepted Answer

The Answer is The superscript gives

A = N + Z ext {. The value of } Z ext { is found in the periodic table for }

these elements to be

\begin{array} { l } Z ( \mathrm { Si } ) = 14 \Z ( \mathrm { Sc } ) = 21 \Z ( \mathrm {~V} ) = 23 \Z ( \mathrm { Bi } ) = 83 \Z ( \mathrm { Pu } ) = 94\end{array}

Since

Z

gives both the number of electrons and protons for the neutral atom, and

N = A - Z

then we find

\begin{array}{llll}\hline ext { nuclide } & ext { neutrons }(\mathrm{N}) & ext { protons }(\mathrm{Z}) & ext { electrons } \\hline{ }^{29} \mathrm{Si} & 15 & 14 & 14 \{ }^{44} \mathrm{Sc} & 23 & 21 & 21 \{ }^{49} \mathrm{~V} & 26 & 23 & 23 \{ }^{210} \mathrm{Bi} & 127 & 83 & 83 \{ }^{241} \mathrm{Pu} & 147 & 94 & 94 \\hline\end{array}

Tabulate the Number of Neutrons, Protons, and Electrons for Neutral