Question

Write the beta decay equation for the following isotope: 15/6 C?

Answer 1

Answer:

$\rm _{6}^{15}\text{C} \longrightarrow \, _{-1}^{0}\text{e} + \, _{7}^{15}\text{N}$

Explanation:

The unbalanced nuclear equation is

$\rm _{6}^{15}\text{C} \longrightarrow \, _{-1}^{0}\text{e} + \, ?$

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

$\rm _{6}^{15}\text{C} \longrightarrow \, _{-1}^{0}\text{e} + \, _{Z}^{A}\text{X}$

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

15 =  0 + A, so A = 15 -   0 = 15, and

 6 = -1 + Z, so Z  =   6 + 1 =    7

Then, your nuclear equation becomes

$\rm _{6}^{15}\text{C} \longrightarrow \, _{-1}^{0}\text{e} + \, _{7}^{15}\text{X}$

Element 7 is nitrogen, so the balanced nuclear equation is

$\rm _{6}^{15}\text{C} \longrightarrow \, _{-1}^{0}\text{e} + \, _{7}^{15}\text{N}$