Question

nguyên tố R có khối là 39, notron nhiều hơn proton 1 hạt . vậy số e ngoài võ là

Answer 1

\(\left\{{}\begin{matrix}P+N=39\\P=E=Z\\N-P=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}P+N=39\\N=P+1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}P+P+1=39\\N=P+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}P=E=Z=19\\N=20\end{matrix}\right.\)

Answer 2

\(\left\{{}\begin{matrix}P+N=39\\P=E=Z\\N-P=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}P+N=39\\N=P+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}N+P+1=39\\N=P+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}P=E=Z=19\\N=20\end{matrix}\right.\)