[ hóa học 8] nguyên tử

M

mylinh115

I

iceghost

Theo đề bài ta có :$p+e+n=2p+n=40 \; (1) \\
(p+e)-n=2p-n=12 \; (2)$
$(1)+(2)=4p=52 \iff p=e=13 \\
(1)-(2)=2n=28 \iff n=14$
 
S

sonsuboy

Theo bài ra,ta có:$p+e+n=40$
Do $p=e$ \Rightarrow $2p+n=40$
$2p-n=12$
\Rightarrow $2p+n+2p-n=40+12$
\Rightarrow $4p=52$ \Rightarrow $p=e=13$
\Rightarrow $n=14$
 
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