Question

The radioactive atom 61/27 co emits a beta particle. write an equation showing the decay

Answer 1

Answer:

$\rm _{27}^{60}\text{Co} \longrightarrow \, _{-1}^{0}\text{e} + \, _{28}^{60}\text{Ni}$

Explanation:

The unbalanced nuclear equation is

$\rm _{27}^{60}\text{Co} \longrightarrow \, _{-1}^{0}\text{e} + \, ?$

Let's write the question mark as a nuclear symbol.

$\rm _{27}^{60}\text{Co} \longrightarrow \, _{-1}^{0}\text{e} + \, _{Z}^{A}\text{X}$

The main point to remember in balancing nuclear equations is that the sums of the superscripts and the subscripts must be the same on each side of the equation.  

Then

60 = 0 + A, so A = 60 - 0 = 60, and

27 = -1 + Z, so Z  = 27 + 1 = 28

Your nuclear equation becomes

$\rm _{27}^{60}\text{Co} \longrightarrow \, _{-1}^{0}\text{e} + \, _{28}^{60}\text{X}$

Element 28 is nickel, so the balanced nuclear equation is

$\rm _{27}^{60}\text{Co} \longrightarrow \, _{-1}^{0}\text{e} + \, _{28}^{60}\text{Ni}$