How would you write a balanced nuclear equations for the emission of an alpha particle by polonium-209 and for the decay of calcium-45 to scandium-45?

1 Answer
Dec 5, 2015

Answer:

#""_84^209"Po" -> ""_82^205"Pb" + ""_2^4alpha#

#""_20^45"Ca" -> ""_21^45"Sc" + ""_text(-1)^0"e" + ""_0^0nu#

Explanation:

The alpha decay of polonum-209, which consists of the emission of an alpha particle, will leave behind the atom of a different element.

Notice that the identity of the element changes after the decay. This means that you're dealing with a nuclear transmutation, a process that converts an atom of a given element into an atom of a different element.

Now, an alpha particle consists of

  • two protons
  • two neutrons

In essence, an alpha particle is a helium-4 nucleus, which means that you can see it as having anatomic number equal to #2# and a mass number equal to #4#.

http://scienceblogs.com/startswithabang/2012/12/12/why-the-world-will-run-out-of-helium/

When a nucleus emits an alpha particle, its atomic number will decrease by #2#, since it's losing two protons, and its mass number will decrease by #4#, since it's losing two protons and two neutrons.

You can thus say that

#""_84^209"Po" -> ""_Z^A"X" + ""_2^4alpha#

Balance the nuclear equation by balancing the atomic number and the mass number for both sides of the equation

#209 = A + 4 " "# and #" "84 = Z + 2#

The atomic number and the mass number of the resulting element will be

#A = 209 - 4 = 205" "# and #" "Z = 84 - 2 = 82#

A quick look in the periodic table will show that you're dealing with lead-205

#""_84^209"Po" -> ""_82^205"Pb" + ""_2^4alpha#

Now for the decay of calcium-45 to scandium-45. Two important things to notice here

  • you're once again dealing with a nuclear transmutation
  • the mass number is conserved following the decay

Turn to the periodic table again and make a note of the atomic numbers of the two elements

#Z_(Ca) = 20 " "# and #" "Z_(Sc) = 21#

This means that you're looking for a nuclear decay that will increase the atomic number by #1# and keep the mass number constant.

As you know, in beta decay, a neutron is being converted into a proton and a beta particle and an antineutrino are being emitted from the nucleus.

A beta particle is simply an electron #beta = ""_text(-1)^0"e"#

This means that you will have

#""_20^45"Ca" -> ""_21^45"Sc" + ""_text(-1)^0"e" + ""_0^0nu#