# Question #e9bc7

##### 1 Answer

#### Explanation:

The thing to remember about nuclear equations is that they must **conserve mass and charge**.

In other words, the overall **mass number** and the overall **atomic number** must be *equal* on both sides of the equation.

In your case, you're dealing with a **fusion reaction** in which two helium-4 nuclides, **alpha particles**, fuse to form a new element, beryllium, **atomic number** of the two elements, i.e. of helium and of beryllium.

You will find

#"For He: " Z = 2#

#"For Be: " Z = 4#

Now, in *isotope notation*, the **atomic number** is added to the bottom-left and the **mass number** is added to the top-left of the chemical symbol.

You will thus have

#""_2^4"He" -># isotope notation for helium-4

#""_4^A"Be" -># isotope notation for the resulting isotope

Your goal now is to find the value of **mass number** of the beryllium isotope. You can now write

#""_color(blue)(2)^color(orange)(4)"He" + ""_color(blue)(2)^color(orange)(4)"He" -> ""_color(blue)(4)^color(orange)(A)"Be"#

Notice that you have

#color(blue)(2) + color(blue)(2) = color(blue)(4) -># charge is conserved

You also need to have

#color(orange)(4) + color(orange)(4) = color(orange)(A) -># mass must be conserved as well

The value of

#4 + 4 = A implies A = 8#

Therefore, the resulting isotope is beryllium-8 and the balanced nuclear equation that describes this fusion reaction looks like this

#""_2^4"He" + ""_2^4"He" -> ""_4^8"Be"#

As a side note, notice what happens when the beryllium-8 isotope fuses with another alpha particle

#""_ 4^8"Be" + ""_ 2^4"He" -> ""_ (color(white)(1)6)^12"C"#

The resulting nuclide is carbon-12, the most abundant isotope of carbon.