Unit 3 - Nuclear Chemistry

Radioisotopes are isotopes of an element that have an unstable nucleus. Due to this instability, the nucleus undergoes a process called decay, where it spontaneously breaks down and transforms into a more stable configuration. This decay can occur through various processes such as alpha decay, beta decay, or gamma decay. Therefore, the correct answer is "decay."

Radioisotopes have unstable nuclei, meaning that they have an imbalance of protons and neutrons. This instability causes the nucleus to undergo spontaneous decay, where it emits radiation in the form of alpha particles, beta particles, or gamma rays. This decay process helps to stabilize the nucleus by reducing the excess energy or changing the number of protons and neutrons. Therefore, the correct answer is decay.

The given nuclear equation represents a decay process where an alpha particle is emitted. In this process, a radioactive nucleus undergoes alpha decay, which involves the emission of an alpha particle. An alpha particle consists of two protons and two neutrons, so the decay process results in the nucleus losing two protons and two neutrons, thereby forming a new element.

Transmutation is the correct answer because it refers to the process of converting an atom of one element into an atom of a different element. This can occur through various nuclear reactions, such as radioactive decay or nuclear fission. Decomposition, saponification, and neutralization do not involve the conversion of one element into another, making them incorrect options for this question.

In a fission reaction, heavy nuclei split into lighter nuclei. This process releases a large amount of energy. Fission reactions are commonly used in nuclear power plants and nuclear weapons, where the splitting of heavy atomic nuclei, such as uranium or plutonium, results in the release of energy in the form of heat and radiation. This reaction is also accompanied by the release of additional neutrons, which can trigger a chain reaction, leading to the release of even more energy.

Nuclear fusion is the correct answer because it involves the combination of two lighter nuclei to form one heavier nucleus. This process releases a significant amount of energy and is the fundamental process that powers the sun and other stars. In nuclear fusion, the nuclei must overcome the strong electrostatic repulsion between them in order to come close enough for the strong nuclear force to bind them together. This reaction is the opposite of nuclear fission, where a heavy nucleus splits into two lighter nuclei.

The correct answer is "alpha particle, beta particle, gamma ray". This is because alpha particles have the least penetrating power among the three options. They are large and heavy, consisting of two protons and two neutrons, and are easily stopped by a sheet of paper or a few centimeters of air. Beta particles, on the other hand, are smaller and have a greater penetrating power. They are high-energy electrons or positrons and can penetrate through a few millimeters of aluminum or several centimeters of air. Gamma rays have the greatest penetrating power. They are electromagnetic waves with high energy and can penetrate through thick layers of material, requiring thick lead or concrete shielding to stop them.

The correct answer is alpha. In nuclear reactions, alpha particles are represented by the symbol α. They are made up of two protons and two neutrons, giving them a positive charge. Alpha particles are relatively large and heavy compared to other particles, and they can be emitted during radioactive decay or nuclear reactions.

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The question states that radiation is not spontaneously emitted from C-12 nuclei or C-13 nuclei. This implies that C-12 and C-13 nuclei are stable and do not undergo radioactive decay. Therefore, the correct answer is nuclei of C-12 and C-13, only.

In the given reaction, the type of particle represented by X is a neutron. This is because a neutron is an uncharged subatomic particle that has a mass slightly greater than that of a proton. Neutrons are commonly found in the nucleus of an atom and play a crucial role in determining the stability and properties of the nucleus. In the reaction, the presence of a neutron suggests that it is involved in a nuclear process or interaction.

During the fission of Pu-239 atoms, energy is released due to the conversion of matter to energy. This process occurs when the nucleus of the Pu-239 atom is split into two smaller nuclei, releasing a large amount of energy in the form of heat and radiation. This is in accordance with Einstein's mass-energy equivalence principle, which states that mass can be converted into energy and vice versa. In this case, a small amount of matter is converted into a large amount of energy, resulting in the release of a significant amount of energy during fission.

C-14 and C-12 are used to date the remains of once-living organisms because they are isotopes of carbon. C-14 is a radioactive isotope that is constantly being formed in the atmosphere and is absorbed by living organisms through photosynthesis. When an organism dies, the C-14 begins to decay at a known rate, allowing scientists to measure the amount of C-14 remaining in the remains and determine how long ago the organism died. C-12, on the other hand, is a stable isotope of carbon that does not decay and is used as a reference to determine the initial amount of C-14 in the remains.

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An alpha particle consists of two protons and two neutrons, making it relatively large and heavy compared to a beta particle. A beta particle is much smaller and lighter, typically an electron or positron. Due to its smaller mass, a beta particle can travel at higher speeds and penetrate materials more easily than an alpha particle. Therefore, it has less mass and greater penetrating power compared to an alpha particle.

Ba-146 is the correct answer because the question asks for the particle represented by X, and Ba-146 is the only option that matches this criteria. The other options (Ba-150, Ca-146, Ba-145) do not match the given particle.

I-131 is used for diagnosing thyroid disorders because it is a radioactive isotope of iodine. It emits gamma rays that can be detected by a special camera or scanner, allowing doctors to visualize the size, shape, and function of the thyroid gland. I-131 is specifically taken up by the thyroid gland, making it a useful tool for diagnosing conditions such as hyperthyroidism and thyroid cancer.

A positron is represented by X in the given nuclear equation. A positron is the antiparticle of an electron, carrying a positive charge instead of a negative charge. In a nuclear equation, a positron is often emitted during certain types of radioactive decay processes, such as beta-plus decay.

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Co-60 is the correct answer because it is a radioactive isotope of cobalt that emits high-energy gamma rays. These gamma rays can penetrate tissues deeply and are used in radiation therapy to kill cancer cells. Co-60 is commonly used in external beam radiation therapy machines, known as cobalt machines, to deliver precise and targeted radiation doses to cancerous tumors. This helps to shrink tumors, destroy cancer cells, and prevent their growth.

The particle represented by X is a proton. Protons are subatomic particles with a positive charge and are found in the nucleus of an atom. They have a mass of approximately 1 atomic mass unit and are essential for the stability and identity of an atom.

Francium-220 is the correct answer because it has a half-life of less than one minute. Half-life is the time it takes for half of the atoms in a radioactive substance to decay. Since francium-220 has a half-life less than one minute, it means that it decays at a faster rate compared to the other nuclides listed. Cesium-137, phosphorus-32, and strontium-90 all have longer half-lives, indicating slower rates of decay.

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U-238 is used to date geological formations like gneiss because it undergoes radioactive decay over time, transforming into Pb-206. By measuring the ratio of U-238 to Pb-206 in a sample, scientists can determine the age of the formation. This method, known as radiometric dating, relies on the predictable rate of decay of U-238 and is commonly used to determine the ages of rocks and minerals in the Earth's crust.

This equation shows that P-32 (phosphorus-32) decays into S-32 (sulfur-32) by emitting a beta particle (β⁻) and an electron antineutrino (ν̅_e).

If you can provide more information about the image, such as the source or the context in which the equations appear, I may be able to help you identify the equation that represents the decay of P-32. 

The half-life of a radioisotope is the time it takes for half of the original sample to decay. In this case, we know that 25.0 grams of the original 100-gram sample remains unchanged after 11.46 days. Since half of the sample remains unchanged, we can assume that the other half has decayed. Therefore, it took 11.46 days for half of the sample to decay. To find the half-life, we divide this time by 2, which gives us 5.73 days. Therefore, the correct answer is 5.73 d.

Use Table N.

The correct answer is 17190 y. This is because C-14 has a half-life of 5730 years, meaning that after 5730 years, half of the original sample will remain. To find the total number of years it takes for only 12.50 grams to remain, we need to multiply the half-life by 3, since 12.50 grams is one-eighth of the original 100.0-gram sample. Therefore, 5730 years x 3 = 17190 years.

Ca-37 and Ne-19 have the same decay mode because they both undergo beta decay. In beta decay, a neutron in the nucleus is converted into a proton, and a high-energy electron (beta particle) is emitted. This process occurs in both Ca-37 and Ne-19, making them the correct answer to the question.

In this question, we are asked to determine the reason why atoms of K-37 spontaneously decay. The correct answer is "unstable nuclei emit positrons". This means that the nuclei of K-37 are not stable and undergo radioactive decay by emitting positrons. Positron emission is a type of radioactive decay where an unstable nucleus emits a positron, which is a positively charged electron. This process helps the unstable nucleus to become more stable.