A) its atomic number is high
B) it has high \[\frac{p}{n}\]ratio
C) it has high\[\frac{n}{p}\]ratio
D) none of the above
Correct Answer: C
Solution :
\[_{\text{27}}^{\text{60}}\text{Co}\]is radioactive and unstable due to high\[\frac{n}{p}\] ratio, \[\left( ie,\frac{n}{p}>1 \right)\] Number of protons \[=27\] Number of neutrons\[=33\] \[\therefore \]\[\frac{n}{p}\]for \[_{27}^{60}Co=\frac{33}{27}=1.22\]
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