ta có P+E+N=28 => 2Z+N=28
P+E-N=8 =>2Z-N=8
=>Z=9 , N =10
=> A là nguyên tố Flo (F)
theo đề bài ta có:
\(p+n+e=28\)
mà \(p=e\)
\(\Rightarrow\left\{{}\begin{matrix}2p+n=28\\2p-n=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2n=20\\2p-n=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}n=10\\2p-10=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}n=10\\p=9\end{matrix}\right.\)
vậy \(p=e=9;n=10\)
\(NTK_X=9+10=19\left(đvC\right)\)
\(\Rightarrow X\) là \(F\left(Flo\right)\)