Question #afc3d

1 Answer
Sep 20, 2017

Answer:

Here's what I got.

Explanation:

For starters, grab a Periodic Table and look for potassium, #"K"#, and argon, #"Ar"#. Once you find them, make a note of their respective atomic numbers.

  • #"For K: " Z_"K" = 19#
  • #"For Ar: " Z_"Ar" = 18#

You can now write the isotope notation for your two isotopes.

  • #"potassium-40 " implies " " ""_19^40"K"#
  • #"argon-40 " implies " " ""_18^40"Ar"#

This means that the unbalanced nuclear equation that describes this radioactive decay process will look like this

#""_19^40"K" -> ""_18^40"Ar" + ""_Z^A?#

Your goal here will be to identify the unknown particle, which I labeled "#?#".

As you know, in any nuclear reaction, mass and charge must be conserved. This implies that you can write two equations

#40 = 40 + A -># conservation of mass

#19 = 18 + Z -># conservation of charge

Solve these equations to get

#40 = 40 + A implies A = 0#

#19 = 18 + Z implies Z = 1#

The particle that has #Z = 1# and #A = 0# is the positron, which is the antiparticle of the electron. The positron is also known as a beta-plus particle, #beta^(+)#.

This means that potassium-40 can decay to argon-40 by way of positron emission, or beta-plus decay.

The balanced nuclear equation that describes the positron emission of potassium-40 looks like this

#""_19^40"K" -> ""_18^40"Ar" + ""_1^0beta + nu_e#

Keep in mind that an electron neutrino, #nu_e# is also emitted here.

Now, you can also set up the unbalanced nuclear equation like this

#""_19^40"K"+ ""_Z^A? -> ""_18^40"Ar" #

This time, you have

#40 + A = 40 -># conservation of mass

#19 + Z = 18 -># conservation of charge

Solve the two equations to get

#A = 0" " and " "Z = -1#

In this case, the unknown particle is an electron, or beta particle, #beta^(-)#. You can thus say that potassium-40 can also decay to argon-40 by way of electron capture.

The balanced nuclear equation that describes the electron capture of potassium-40 looks like this

#""_ 19^40"K" + ""_ (-1)^(color(white)(-)0)beta -> ""_18^40"Ar" + nu_e#

Once again, notice that the electron capture results in the emission of an electron neutrino.