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When a target nucleus is bombarded by an appropriate beam of particles, it is possible to produce a) a less massive nucleus, but not a more massive one. b) a more massive nucleus, but not a less massive one. c) a nucleus with smaller charge number, but not one with a greater charge number. d) a nucleus with greater charge number, but not one with a smaller charge number. e) a nucleus with either greater or smaller charge number.

Short Answer

Expert verified
Answer: e) A nucleus with either greater or smaller charge number can be produced when a target nucleus is bombarded by an appropriate beam of particles.

Step by step solution

01

Restate the exercise

Determine which statement is correct when a target nucleus is bombarded by an appropriate beam of particles.
02

Consider fusion and fission reactions

When a target nucleus is bombarded by particles, it can experience either nuclear fusion or nuclear fission. Fusion is the process where two lighter nuclei combine to form a more massive nucleus, while fission is the process where a heavier nucleus splits into two or more lighter nuclei.
03

Analyze each statement

a) Producing a less massive nucleus is possible through fission but not through fusion. However, a more massive nucleus can be produced through fusion, so this statement is incorrect. b) Producing a more massive nucleus is possible through fusion but not through fission. However, a less massive nucleus can be produced through fission, so this statement is also incorrect. c) In both fusion and fission reactions, the charge number can either increase or decrease, so this statement is incorrect. d) In both fusion and fission reactions, the charge number can either increase or decrease, so this statement is incorrect as well. e) In both fusion and fission reactions, the charge number can either increase or decrease, depending on the specific reaction taking place. So, this statement is correct.
04

Answer

The correct statement is e) a nucleus with either greater or smaller charge number can be produced when a target nucleus is bombarded by an appropriate beam of particles.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Nuclear Fusion
Nuclear fusion is a fundamental concept in nuclear physics. It refers to a reaction where two light atomic nuclei come together to form a heavier nucleus. This process releases a significant amount of energy. Fusion is actually what powers the sun and other stars.
Scientists have long been interested in harnessing this energy for practical uses here on Earth. Here are some important things to keep in mind about nuclear fusion:
  • Fusion reactions require very high temperatures and pressures to overcome the repulsive forces between the positively charged nuclei.
  • When fusion happens, it leads to the formation of a new nucleus that is heavier than either of the starting nuclei.
  • Fusion is considered a cleaner source of energy compared to nuclear fission, as it produces less radioactive waste.
Despite its potential, achieving controlled nuclear fusion on Earth is extremely challenging and remains a topic of ongoing research.
Breaking Down Nuclear Fission
Nuclear fission is another crucial type of nuclear reaction. In fission, a heavy atomic nucleus splits into two or more smaller nuclei. This process also releases energy, which is used in nuclear power plants to generate electricity.
In addition to energy production, fission has other important attributes:
  • Fission usually involves bombarding a heavy nucleus, like Uranium-235, with a neutron, which causes it to split.
  • The splitting of the nucleus results in the emission of additional neutrons and other smaller particles, along with a large amount of energy.
  • The emitted neutrons can further induce fission in other nuclei, leading to a chain reaction.
  • Unlike fusion, fission reactions can be controlled more easily, but they produce radioactive byproducts.
Understanding fission is key to comprehending how current nuclear reactors function and why managing nuclear waste is important.
Nuclear Charge Number
The nuclear charge number is an essential property of an atomic nucleus. It is equivalent to the atomic number and represents the number of protons in the nucleus. This number defines the identity of an element, distinguishing it from other elements in the periodic table.
Here are some aspects related to nuclear charge number:
  • In nuclear reactions such as fission and fusion, the nuclear charge number can change, resulting in the transformation of one element into another.
  • A higher nuclear charge number means more protons, contributing to a greater positive charge and stronger electromagnetic forces.
  • Changes in nuclear charge number are pivotal in understanding nuclear transmutation processes.
Keeping track of the nuclear charge number helps scientists predict and categorize the outcomes of nuclear reactions, as seen in various applications, including energy production and medical therapies.

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Most popular questions from this chapter

The specific activity of a radioactive material is the number of disintegrations per second per gram of radioactive atoms. a) Given the half-life of \({ }^{14} \mathrm{C}\) of \(5730 \mathrm{yr}\), calculate the specific activity of \({ }^{14} \mathrm{C}\). Express your result in disintegrations per second per gram, becquerel per gram, and curie per gram. b) Calculate the initial activity of a \(5.00-\mathrm{g}\) piece of wood. c) How many \({ }^{14} \mathrm{C}\) disintegrations have occurred in a \(5.00-\mathrm{g}\) piece of wood that was cut from a tree January \(1,1700 ?\)

The most common isotope of uranium, \({ }_{92}^{238} \mathrm{U},\) produces radon \({ }_{86}^{222} \mathrm{Rn}\) through the following sequence of decays: $$\begin{array}{c}{ }^{238} \mathrm{U} \rightarrow{ }^{234} \mathrm{Th}+\alpha,{ }^{234} \mathrm{Th} \rightarrow{ }^{234} \mathrm{~Pa}+\beta^{-}+\bar{\nu}_{e}, \\\\{ }_{91}^{234} \mathrm{~Pa} \rightarrow{ }_{92}^{234} \mathrm{U}+\beta+\bar{\nu}_{e},{ }^{234} \mathrm{U} \rightarrow{ }^{230} \mathrm{Th}+\alpha ,\\\\{ }_{91}^{230} \mathrm{Th} \rightarrow{ }_{90}^{226} \mathrm{Ra}+\alpha,{ }_{88}^{226} \mathrm{Ra} \rightarrow{ }_{86}^{222} \mathrm{Rn}+\alpha,\end{array}$$. A sample of \({ }_{92}^{238} \mathrm{U}\) will build up equilibrium concentrations of its daughter nuclei down to \({ }_{88}^{226} \mathrm{Ra} ;\) the concentrations of each are such that each daughter is produced as fast as it decays. The \({ }_{88}^{226} \mathrm{Ra}\) decays to \({ }_{86}^{222} \mathrm{Rn},\) which escapes as a gas. (The \(\alpha\) particles also escape, as helium; this is a source of much of the helium found on Earth.) In high concentrations, the radon is a health hazard in buildings built on soil or foundations containing uranium ores, as it can be inhaled. a) Look up the necessary data, and calculate the rate at which \(1.00 \mathrm{~kg}\) of an equilibrium mixture of \({ }_{92}^{238} \mathrm{U}\) and its first five daughters produces \({ }_{86}^{222} \mathrm{Rn}\) (mass per unit time). b) What activity (in curies per unit time) of radon does this represent?

The mass of an atom (atomic mass) is equal to a) the sum of the masses of the protons. b) the sum of the masses of protons and neutrons. c) the sum of the masses of protons, neutrons and electrons. d) the sum of the masses of protons, neutrons, and electrons minus the atom's binding energy.

Refer to the subsection "Terrestrial Fusion" in Section 40.4 to see how achieving controlled fusion would be the solution to mankind's energy problems, and how difficult it is to do. Why is it so hard? The Sun does it all the time (see the previous subsection, "Stellar Fusion"). Do we need to understand better how the Sun works to build a useful nuclear fusion reactor?

A certain radioactive isotope decays to one-eighth its original amount in \(5.0 \mathrm{~h}\). a) What is its half-life? b) What is its mean lifetime?

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