# How is radiactive dating performed dating your definition of cougar

(Note that this does not mean that the ratios are the same everywhere on earth.It merely means that the ratios are the same in the particular magma from which the test sample was later taken.) As strontium-87 forms, its ratio to strontium-86 will increase.The sum of protons plus neutrons is the mass number.We designate a specific group of atoms by using the term "nuclide." A nuclide refers to a group of atoms with specified atomic number and mass number.The amount of strontium-86 in a given mineral sample will not change.Therefore the relative amounts of rubidium-87 and strontium-87 can be determined by expressing their ratios to strontium-86: Rb-87/Sr-86 and Sr87/Sr-86 We measure the amounts of rubidium-87 and strontium-87 as ratios to an unchanging content of strontium-86.If we knew the fraction of a radioactive element still remaining in a mineral, it would be a simple matter to calculate its age by the formula To determine the fraction still remaining, we must know both the amount now present and also the amount present when the mineral was formed.

The number of protons in the nucleus of an atom is called its atomic number.Because of radioactivity, the fraction of rubidium-87 decreases from an initial value of 100% at the time of formation of the mineral, and approaches zero with increasing number of half lives.At the same time, the fraction of strontium-87 increases from zero and approaches 100% with increasing number of half-lives.Radioactive elements "decay" (that is, change into other elements) by "half lives." If a half life is equal to one year, then one half of the radioactive element will have decayed in the first year after the mineral was formed; one half of the remainder will decay in the next year (leaving one-fourth remaining), and so forth.The formula for the fraction remaining is one-half raised to the power given by the number of years divided by the half-life (in other words raised to a power equal to the number of half-lives).

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