Ta có \(1\le\frac{N}{Z}\le1,52\)
\(\Rightarrow\frac{2Z+N}{3,52}\le Z\le\frac{2Z+N}{3}\)
\(\Leftrightarrow\frac{52}{3,52}\le Z\le\frac{52}{3}\)
\(\Leftrightarrow14,77\le Z\le17,33\)
\(\Rightarrow\left[{}\begin{matrix}Z=15\\Z=16\\Z=17\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}N=22\left(loại\right)\\N=20\left(loại\right)\\N=18\left(chọn\right)\end{matrix}\right.\)
\(\Rightarrow Z=E=P=17,N=18\)
\(\Rightarrow\)X là Cl
Theo bài ra 2p + n = 52 <=> n = 52-2p
Ta có \(1\le\frac{n}{p}\le1,52\) <=> \(1\le\frac{52-2p}{p}\le1,52\) <=> 14,77\(\le p\le17,33\)
| p | 15 | 16 | 17 |
| n | 22(l) | 20(l) | 18(t.m) |
Vậy p=e=17 (hạt ) , n = 18 hạt
