# What is the radioactive particle released in the following nuclear equation #""_74^159"W" -> ""_72^155"Hf" + ? #

##### 1 Answer

#### Answer:

#### Explanation:

The nuclear equation given to you describes the decay of *tungsten-159* to *hafnium-155*

#""_ (color(white)(a)74)^159"W" -> ""_ (color(white)(a)72)^155"Hf" + ""_Z^A?#

Your goal here is to find the values of

Here **mass number** of the particle, which as you know tells you the number of protons * and* electrons present in the nucleus, and

**atomic number**, which is simply the number of protons present in the nucleus.

In any nuclear reaction, **charge** and **mass** must be conserved. This means that you can write

#159 = 155 + A -># conservation of mass

#color(white)(a)74 = color(white)(a)72 + Z -># conservation of charge

Solve these two equations to find

#A = 159 - 155 = 4#

#Z = 74 - 72 =2#

You can thus say that the unknown particle will have a mass number equal to **alpha particle**, *nucleus* of a helium-4 atom.

The balanced nuclear equation will thus look like this

#color(green)(|bar(ul(color(white)(a/a)color(black)(""_ (color(white)(a)74)^159"W" -> ""_ (color(white)(a)72)^155"Hf" + ""_2^4alpha)color(white)(a/a)|)))#