# Balance the following nuclear equations ""_(10)^(24) "Ne" -> ??? + beta

##### 1 Answer
Nov 21, 2017

""_10^24"Ne" → ""_11^24"Na" + β

#### Explanation:

Your equation is

""_10^24"Ne" → ??? + β

As you might guess this is an example of β-decay.

Recall that a β-particle is just an electron.

It becomes easier to balance the equation if we use an alternate symbol for a β-particle: $\text{_text(-1)^0"e}$.

Then the equation becomes

$\text{_10^24"Ne" → ??? +""_text(-1)^0"e}$

Now, let's replace the ??? with an element symbol $\text{_x^y"Z}$. then the equation becomes

$\text{_10^24"Ne" → ""_x^y"Z" +""_text(-1)^0"e}$

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

Sum of superscripts: $24 = y + 0$, so $y = 24$

Sum of subscripts: $\textcolor{w h i t e}{m} 10 = x - 1$, so $x = 11$.

This gives us

$\text{_10^24"Ne" → ""_11^24"Z" +""_text(-1)^0"e}$

The atomic number of the element is 11, and element 11 is sodium.

∴ The nuclear equation is

$\text{_10^24"Ne" → ""_11^24"Na" + ""_text(-1)^0"e}$

If desired, we can now re-insert the β symbol.

""_10^24"Ne" → ""_11^24"Na" + β