Chapter 39, Problem 64A

Cobalt-60 undergoes beta decay. Show that the beta decay of cobalt-60 $\left({}_{27}^{60}Co\right)$ results in nickel-60 $\left({}_{28}^{60}Ni\right)$.

To determine

Proof for Nickel due to beta decay of Cobalt.

Explanation of Solution

Introduction:

The number of proton in the nucleus of new element increase by one unit due to decay of one beta particle from any element. The atomic number is same as the number of protons in the nucleus of element and atomic mass is the sum of number of protons and neutrons in the nucleus of the element.

The atomic number of the ${}_{27}^{60}\text{Co}$ is 27 and atomic mass number is 60. The number of protons in the nucleus of Cobalt is also 27. The number of protons in the new element by the decay of one beta particle from Cobalt is 28.

The expression for the beta decay from Cobalt is,

${}_{27}^{60}\text{Co}{\to }_{28}^{60}\text{Ni}{+}_{-1}^0\text{β}$

The new element is Nickel because the number of protons in the nucleus of Nickel is 28.

Conclusion:

Thus, the new element is nickel because the number of protons in the nucleus of nickel is 28.