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The most important thing to know is that an α-particle (alpha particle) is a helium nucleus.
It contains 2 protons and 2 neutrons, for a mass number of 4.
During α-decay, an atomic nucleus emits an alpha particle. It transforms (or decays) into an atom with an atomic number 2 less and a mass number 4 less.
Thus, radium-226 decays through α-particle emission to form radon-222 according to the equation:
##””_88^226″Ra”## → ##””_86^222″Rn” + _2^4″He”##
Note that the sum of the subscripts (atomic numbers or charges) is the same on each side of the equation. Also, the sum of the superscripts (masses) is the same on each side of the equation.
EXAMPLE
Write a balanced nuclear equation for the α decay of polonium-208.
Solution
The unbalanced equation is
##””_84^208″Po”## → ##””_2^4″He” + “X”##
The superscript of ##”X”## must be 208 – 4 = 204.
The subscript of ##”X”## must be 84 – 2 = 82.
Element 82 is Pb. So the equation is
##””_84^208″Po”## → ##””_2^4″He” + _82^204″Pb”##
Here is a video that describes how to write equations for .
Hope this helps!
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