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 :
\[_{27}^{60}Co\]is radioactive and unstable due to high\[\frac{n}{p}\]ratio, \[\left( i.e.,,\frac{n}{p}>1 \right)\] Number of proton = 27 Number of neutron =33 \[\therefore \] \[\frac{n}{p}\]from \[_{27}^{60}Co=\frac{33}{27}=1.22\]
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