Question

The isotope of cobalt ${}_{27}^{60}Co$ undergoes $\beta$-decay according to the given equation
${}_{27}^{60}Co{\to }_z^{A}X{+}_{-1}^{0}e$.
Find out the element X from the above nuclear fission reaction?

• A. ${}_{26}^{60}Fe$
• B. ${}_{25}^{56}Mn$
• C. ${}_{28}^{60}Ni$
• D. ${}_{29}^{60}Cu$
• E. ${}_{27}^{61}Co$

Answer 1

The correct option is C ${}_{28}^{60}Ni$

In a nuclear reaction, the atomic number is conserved as well as the mass number is conserved.

$\therefore$ $60=A+0$ $⟹A=60$

Also, $27=Z+(-1)$ $⟹Z=28$

Thus the element $X$ is ${}_{28}^{60}Ni$.