Question #5dc62

All you really need to use in order to answer this question is a Periodic Table and the fact that when a radioactive nuclide undergoes alpha decay

• its atomic number decreases by two, #color(green)(Z - 2)#
• its mass number decreases by four, #color(blue)(A - 4)#

This is the case because when a radioactive nuclide undergoes alpha decay, its emits an alpha particle, which is essentially the nucleus of a helium-4 atom.

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So, grab a Periodic Table and look for astatine, #"At"#. You'll find it located in period 6, group 17.

Astatine has an atomic number equal to #color(green)(85)#, which means that you can write the starting nuclide as

#""_(color(white)(1)color(green)(85))^color(blue)(218)"At"#

Since an alpha particle contains #2# protons and #2# neutrons, you can say that

#""_ (color(white)(1)color(green)(85))^color(blue)(218)"At" -> ""_ color(green)(Z)^color(blue)(A)"?" + ""_ color(green)(2)^color(blue)(4)alpha#

Your goal here is to find the identity of the unknown nuclide, which we labeled as "#?#".

As you know, mass and charge must be conserved in a nuclear reaction. This implies that the total mass number and the total atomic number must be equal on both sides of the equation.

You can thus say that

• conservation of mass # ->" "color(blue)(218 = A + 4)#

• conservation of charge # -> " "color(green)(85 = Z + 2)#

This will get you

#color(blue)(A = 218 - 4 = 214)" "# and #" "color(green)(Z = 85 - 2 = 83)#

Check the Periodic Table for the element that has an atomic number equal to #83#. You'll find that bismuth is the match, so

#""_ color(green)(Z)^color(blue)(A)"?" = ""_ (color(white)(1)color(green)(83))^color(blue)(214)"Bi"#

Therefore, you can say that the alpha decay of astatine-218 will produce bismuth-214, or #""^214"Bi"#.

The complete nuclear equation will be

#""_ (color(white)(1)color(green)(85))^color(blue)(218)"At" -> ""_ (color(white)(1)color(green)(83))^color(blue)(214)"Bi" + ""_ color(green)(2)^color(blue)(4)alpha#