# Maximum dating age equation osu dating

Example $$\Page Index$$ If there are 60 grams of $$\ce$$-240 present, how much $$\ce$$-240 will remain after 4 hours?

($$\ce$$-240 has a half-life of 1 hour) Solution $\text = \dfrac (60 \, \text) \nonumber$ After 4 hours, only $$3.75 \: \text$$ of our original $$60 \: \text$$ sample would remain the radioactive isotope $$\ce$$-240.

Each radioactive isotope will have its own unique half-life that is independent of any of these factors.

Figure $$\Page Index$$: For cobalt-60, which has a half-life of 5.27 years, 50% remains after 5.27 years (one half-life), 25% remains after 10.54 years (two half-lives), 12.5% remains after 15.81 years (three half-lives), and so on. The half-lives of many radioactive isotopes have been determined and they have been found to range from extremely long half-lives of 10 billion years to extremely short half-lives of fractions of a second.

Figure $$\Page Index$$: Along with stable carbon-12, radioactive carbon-14 is taken in by plants and animals, and remains at a constant level within them while they are alive.

After death, the C-14 decays and the C-14: C-12 ratio in the remains decreases. For example, a sample can be C-14 dating if it is approximately 100 to 50,000 years old.

$\text = \dfrac (\text)$ where $$n$$ is the number of half-lives.A useful concept is half-life (symbol is $$t_$$), which is the time required for half of the starting material to change or decay.Half-lives can be calculated from measurements on the change in mass of a nuclide and the time it takes to occur.$720\cancel\times \dfrac= 30\, days \nonumber$ $n=3 =\dfrac \nonumber$ $\text = \dfrac (8.0 \, ug) \nonumber$ After 720 hours, 1.0 ug of the material remains as $$\ce$$-225 .For example, carbon-14 has a half-life of 5,730 years and is used to measure the age of organic material.

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