\(\left\{{}\begin{matrix}P=E\\P+E+N=40\\N=\dfrac{7}{13}\left(P+E\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}P=E=13\\N=14\end{matrix}\right.\\ A=M=P+N=13+14=27\)
ta có :2p+n=40
->\(\left\{{}\begin{matrix}2p+n=40\\n=\dfrac{7}{13}2p\end{matrix}\right.\)
->p=e=13 hạt
=>n=40-(13.2)=14 hạt
=>A=27
->Z là nhôm (Al)