A) \[_{28}^{59}Ni\]
B) \[_{27}^{59}Co\]
C) \[_{28}^{60}Ni\]
D) \[_{27}^{60}Co\]
Correct Answer: C
Solution :
\[_{28}N{{i}^{60}}+\underset{(n)}{\mathop{_{0}{{n}^{1}}}}\,{{\xrightarrow[{}]{{}}}_{27}}C{{o}^{60}}+\underset{(p)}{\mathop{_{1}{{H}^{1}}}}\,\] (n, p means that neutron attacks and proton liberates.)
You need to login to perform this action.
You will be redirected in
3 sec
