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?
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
When a nucleus emits an alpha particle, its atomic number will decrease by
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
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
This means that you will have
#""_20^45"Ca" -> ""_21^45"Sc" + ""_text(-1)^0"e" + ""_0^0nu#