Untitled - Kelly Walsh High School

Nuclear Chemistry 293
The sum of the mass numbers on the left of the equation is 36 (35 1) and on
the right is 1 x. The mass number of the unknown isotope must be 35. The
sum of the atomic numbers on the left is 17 (17 0) and 1 y on the right. The
atomic number of the unknown must then be 16. This atomic number identifies
the element as sulfur, so that we can write a complete nuclear equation:
Sulfur-35 does not occur in nature; it is an artificial or man-made isotope.
l
We can observe three common types of radioactive decay in nature. We can
occasionally observe others.
Alpha Emission
An alpha particle is essentially a helium nucleus with two protons and two neu-
4
trons. It is represented as or 2 . As this particle leaves the decaying nucleus
it has no electrons and thus has a 2 charge. However, it quickly acquires two
electrons from the surroundings to form the neutral atom. Most commonly, we
show the alpha particle as the neutral particle and not the cation.
a
4
2He Radon-222 undergoes alpha decay according to the following balanced equation:
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Notice that in going from Rn-222 to Po-218, the atomic number has decreased
by 2 and the mass number by 4.
Beta Emission
35
17Cl 1 0n 222
86 Rn
A beta particle is essentially an electron and can be represented as either or
0
1 . This electron comes from the nucleus and not the electron cloud. It results
from the conversion of a neutron into a proton and an electron:
e
0
1b 1
0n l
Nickel-63 will undergo beta decay according to the following equation:
63
28Ni l
1 35
1H 16S 218
84 Po 4 2 He
1 0
1p 1e 63 0
29Cu 1e Notice that the atomic number has increased by one in going from nickel-63 to
copper-63 but the mass number has remained unchanged.