Define half-life together it relates to radioactive nuclides and also solve half-life problems. Explain the general process by i m sorry radioactive dating is provided to identify the period of miscellaneous objects. Calculate the time for a sample come decay. Complete dosage calculations based on nuclide activity.

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The rate of radioactive decay is often identified by the half-life the a radioisotope. Half-life \$$\\left( t_1/2 \\right)\$$ is the time compelled for one half of the nuclei in a sample that radioactive material to decay. After every half-life has passed, one fifty percent of the radioactive nuclei will have actually transformed into a new nuclide (see table below). The rate of decay and also the half-life do not rely on the original size that the sample. They also do not rely upon environmental factors such together temperature and also pressure.

Table \$$\\PageIndex1\$$ Number the Half-Lives PassedFraction RemainingPercentage RemainingMass remaining starting with \$$80 \\: \\textg\$$
1 1/2 50 \$$40 \\: \\textg\$$
2 1/4 25 \$$20 \\: \\textg\$$
3 1/8 12.5 \$$10 \\: \\textg\$$
4 1/16 6.25 \$$5.0 \\: \\textg\$$
5 1/32 3.125 \$$2.5 \\: \\textg\$$

As an example, iodine-131 is a radioisotope v a half-life that 8 days. The decays through beta particle emission into xenon-131.

\\<\\ce^131_53I \\rightarrow \\ce^131_54Xe + \\ce^0_-1e\\>

After eight days have passed, half of the atoms of any sample that iodine-131 will have decayed, and the sample will now be \$$50\\%\$$ iodine-131 and also \$$50\\%\$$ xenon-131. After an additional eight days pass (a full of 16 days or 2 half-lives), the sample will be \$$25\\%\$$ iodine-131 and \$$75\\%\$$ xenon-131. This proceeds until the whole sample the iodine-131. Has fully decayed (see number below). Figure \$$\\PageIndex1\$$: The half-life the iodine-131 is eight days. Half of a provided sample of iodine-131 decays after every eight-day time period elapses.

Half-lives have actually a an extremely wide range, indigenous billions of years to fountain of a second. Listed below (see table below) room the half-lives of part common and important radioisotopes. Those with half-lives on the range of hrs or days are the persons most suitable for usage in clinical treatment.

Table \$$\\PageIndex2\$$ NuclideHalf-Life \$$\\left( t_1/2 \\right)\$$Decay Mode
Carbon-14 5730 years \$$\\beta^-\$$
Cobalt-60 5.27 years \$$\\beta^-\$$
Francium-220 27.5 seconds \$$\\alpha\$$
Hydrogen-3 12.26 years \$$\\beta^-\$$
Iodine-131 8.07 days \$$\\beta^-\$$
Nitrogen-16 7.2 seconds \$$\\beta^-\$$
Phosphorus-32 14.3 days \$$\\beta^-\$$
Plutonium-239 24,100 years \$$\\alpha\$$
Potassium-40 \$$1.28 \\times 10^9\$$ years \$$\\beta^-\$$ and \$$\\cee^-\$$ capture
Radium-226 1600 years \$$\\alpha\$$
Radon-222 3.82 days \$$\\alpha\$$
Strontium-90 28.1 days \$$\\beta^-\$$
Technetium-99 \$$2.13 \\times 10^5\$$ years \$$\\beta^-\$$
Thorium-234 24.1 days \$$\\beta^-\$$
Uranium-235 \$$7.04 \\times 10^8\$$ years \$$\\alpha\$$
Uranium-238 \$$4.47 \\times 10^9\$$ years \$$\\alpha\$$

The following example illustrates how to usage the half-life the a sample to identify the quantity of radioisotope that stays after a certain duration of time has passed.

## Decay Series

In plenty of instances, the decay of an unstable radioactive nuclide simply produces one more radioactive nuclide. It may take number of successive procedures to reach a nuclear species that is stable. A decay series is a succession of succeeding radioactive decays that proceeds till a secure nuclide is reached. The terms reactant and product are normally not used for nuclear reactions. Instead, the state parent and also daugher nuclide are provided to to describe the starting and ending isotopes in a decay process. The figure below shows the decay series for uranium-238. Figure \$$\\PageIndex2\$$: The decay of uranium-238 proceeds along countless steps till a steady nuclide, lead-206, is reached. Each degeneration has its very own characteristic half-life.

In the an initial step, uranium-238 decays through alpha emission to thorium-234 v a half-life of \$$4.5 \\times 10^9\$$ years. This reduce its atom number by two. The thorium-234 promptly decays by beta emission to protactinium-234 (\$$t_1/2 =\$$ 24.1 days). The atom number rises by one. This proceeds for many much more steps until eventually the series ends v the development of the secure isotope lead-206.

## Artificial Transmutation

As we have seen, transmutation occurs as soon as atoms the one facet spontaneously decay and are convert to atoms of another element. man-made transmutation is the bombardment of steady nuclei through charged or uncharged corpuscle in order to cause a nuclear reaction. The bombarding particles deserve to be protons, neutrons, alpha particles, or bigger atoms. Ernest Rutherford performed few of the faster bombardments, including the bombardment that nitrogen gas with alpha particles to produce the rough fluorine-18 isotope.

\\<\\ce^14_7N + \\ce^4_2He \\rightarrow \\ce^18_9F\\>

Fluorine-18 easily decays to the steady nuclide oxygen-17 by publication a proton.

\\<\\ce^18_9F \\rightarrow \\ce^17_8O + \\ce^1_1H\\>

When beryllium-9 is bombarded through alpha particles, carbon-12 is created with the release of a neutron.

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\\<\\ce^9_4Be + \\ce^4_2He \\rightarrow \\ce^12_6C + \\ce^1_0n\\>

## Dosing

Half-life calculations can be based on mass, percent remaining, or dose. Regardless of which one, the ide is tho the same. Understanding the radioactivity and also half-life the a sample is crucial for calculating the exactly dose because that a patient and determining the levels and duration the radioactive emissions from a patience after treatment is received.

Frequently, dosages because that radioactive isotopes are offered the activity in volume. Because that example, the concentration of \$$\\ceI\$$-137 is given as \$$50 \\: \\mu \\textCi/mL\$$ (microCurie per milliliter). This relationship deserve to be offered to calculation the volume essential for a details dose. Because that example, a patient needs \$$125 \\: \\mu \\textCi\$$ the \$$\\ceI\$$1-51. What volume that a \$$50 \\mu \\textCi\$$ every \$$10 \\: \\textmL\$$ solution have to be given?