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 protons\[=27\] Number of neutrons\[=33\] \[\therefore \]\[\frac{n}{p}\]for\[_{27}^{60}Co=\frac{33}{27}=1.22\]
You need to login to perform this action.
You will be redirected in
3 sec
