What particle is needed to complete this nuclear reaction? #"_86^222Rn##->##"_84^218 Po + underline# ?

1 Answer
Nov 30, 2016

Answer:

An alpha particle.

Explanation:

The thing to remember about nuclear equations is that mass and charge must always be conserved.

In other words, in any nuclear equation

  • the overall mass number remains unchanged
  • the overall atomic number remains unchanged

Your unbalanced nuclear equation looks like this

#""_ (color(white)(1)color(blue)(86))^color(darkgreen)(222)"Rn" -> ""_ (color(white)(1)color(blue)(84))^color(darkgreen)(218)"Po" + ""_ (color(blue)(Z))^color(darkgreen)(A)"?"#

The goal here is to find the atomic number, #color(blue)(Z)#, and the mass number, #color(darkgreen)(A)#, of the unknown particle.

Since the overall mass number must be conserved, you can say that

#color(darkgreen)(222) = color(darkgreen)(218) + color(darkgreen)(A) -># conservation of mass

This will get you

#color(darkgreen)(A) = 222 - 218 = 4#

The overall atomic number is also conserved, so you can say that

#color(blue)(86) = color(blue)(84) + color(blue)(Z)#

This will get you

#color(blue)(Z) = 86 - 84 = 2#

The unknown particle has a mass number equal to #4# and an atomic number equal to #2#, which means that you're dealing with an alpha particle.

https://www.mirion.com/introduction-to-radiation-safety/types-of-ionizing-radiation/

In essence, an alpha particle is simply the nucleus of a helium-4 atom, i.e. it contains #2# protons and #2# neutrons.

You can now complete the nuclear equation that describes the alpha decay of radon-222 to polonium-218

#""_ (color(white)(1)86)^222"Rn" -> ""_ (color(white)(1)84)^218"Po" + ""_ 2^4alpha#