Problem 11
Question
The element used for carrying out the nuclear reaction is (a) Thorium-232 (b) Uranium-238 (c) Plutonium-239 (d) Neptunium-293
Step-by-Step Solution
Verified Answer
Plutonium-239 (option c) is used for nuclear reactions.
1Step 1: Understanding Nuclear Reactions
Nuclear reactions involve changes in an atom's nucleus and can release a significant amount of energy. Elements used in these reactions are typically isotopes that can undergo fission or fusion. Fission involves the splitting of a heavy nucleus into lighter nuclei, which is commonly utilized in nuclear reactors.
2Step 2: Identify Common Fissionable Materials
Identify isotopes that are commonly used in nuclear reactors or reactions due to their ability to sustain a chain reaction. The most common isotopes are Uranium-235 (not listed here), Uranium-238 (used with a neutron to breed Plutonium-239), and Plutonium-239. These materials can undergo fission after absorption of a neutron.
3Step 3: Analyze Given Options
Examine each option to determine its relevance:
(a) Thorium-232 is used to breed Uranium-233 but is not directly used in fission.
(b) Uranium-238 is not directly fissionable but can be converted into Plutonium-239.
(c) Plutonium-239 can undergo fission, making it a common nuclear fuel.
(d) Neptunium-293 is not commonly used or significant in nuclear reactions.
4Step 4: Choose the Correct Option
From the analysis, Plutonium-239 is the element directly used for fission in nuclear reactions, making option (c) the correct choice.
Key Concepts
FissionIsotopesNuclear Fuel
Fission
Fission is a fundamental principle in nuclear physics, playing a key role in nuclear reactors and atomic energy generation. It involves the splitting of a heavy atomic nucleus, such as uranium or plutonium, into two lighter nuclei. This process releases a substantial amount of energy, often accompanied by the emission of neutrons and gamma radiation. The released neutrons can collide with other atomic nuclei, inducing further fission reactions in a chain reaction. This is the principle behind maintaining a sustained nuclear reaction in reactors.
To understand how fission works, consider an atomic nucleus as a tightly packed bundle of protons and neutrons held together by nuclear forces. When a neutron is absorbed by a large fissile nucleus, the balance of these forces is disrupted. The nucleus becomes unstable and splits into two smaller fragments, releasing energy.
To understand how fission works, consider an atomic nucleus as a tightly packed bundle of protons and neutrons held together by nuclear forces. When a neutron is absorbed by a large fissile nucleus, the balance of these forces is disrupted. The nucleus becomes unstable and splits into two smaller fragments, releasing energy.
- The energy originates from the conversion of mass into energy, as described by Einstein's equation: \[ E = mc^2 \]
- Fissionable isotopes, like Uranium-235 or Plutonium-239, are vital for this process because they readily absorb neutrons.
- Chain reactions require a minimum mass of fissile material, known as critical mass, to maintain the reaction.
Isotopes
Isotopes are atoms of the same element that have the same number of protons but different numbers of neutrons in their nuclei. This variation in neutron number does not affect the chemical behavior of the element but significantly impacts its nuclear properties.
Isotopes are crucial in nuclear reactions because their nuclear stability and ability to undergo fission or fusion depend on the number of neutrons. For instance, Uranium exists primarily in two isotopic forms: Uranium-235 and Uranium-238. While both isotopes share the same chemical properties, only Uranium-235 is readily fissionable under normal conditions.
Isotopes are crucial in nuclear reactions because their nuclear stability and ability to undergo fission or fusion depend on the number of neutrons. For instance, Uranium exists primarily in two isotopic forms: Uranium-235 and Uranium-238. While both isotopes share the same chemical properties, only Uranium-235 is readily fissionable under normal conditions.
- Uranium-238 is more abundant but not directly usable for fission.
- Through neutron absorption and subsequent radioactive decay, Uranium-238 can be transmuted into Plutonium-239, which is fissionable.
- Isotopes like Thorium-232 can also undergo similar processes to produce fissionable materials indirectly.
Nuclear Fuel
Nuclear fuel is the material used in nuclear reactors to sustain a fission chain reaction. The fuel's role is to provide fissile material that can absorb neutrons and maintain ongoing fission reactions.
Common nuclear fuels include Uranium-235 and Plutonium-239, among others. These isotopes are capable of sustaining a chain reaction due to their ability to undergo fission upon neutron absorption. Let's explore some aspects of nuclear fuel:
Common nuclear fuels include Uranium-235 and Plutonium-239, among others. These isotopes are capable of sustaining a chain reaction due to their ability to undergo fission upon neutron absorption. Let's explore some aspects of nuclear fuel:
- Uranium dioxide (UO2) is a typical form of uranium fuel used in many reactors.
- Plutonium can be formed from Uranium-238, making it an important alternative fuel, especially in fast-breeder reactors.
- Thorium cycle involves using Thorium-232 to breed Uranium-233, another potential nuclear fuel.
Other exercises in this chapter
Problem 9
The reactor used to convert stable form to fossile form is (a) Breeder reactor (b) atomic reactor (c) chemical reactor (d) nuclear reactor
View solution Problem 10
The moderator used in nuclear reactors is (a) TEL (b) \(\mathrm{D}_{2} \mathrm{O}\) (c) \(\mathrm{H}_{2} \mathrm{O}_{2}\) (d) \(\mathrm{R}-\mathrm{O}-\mathrm{R}
View solution Problem 13
If the mass defect of a nuclide is \(3.32 \times 10^{-26} \mathrm{~g}\), its binding energy is ..... MeV (a) \(27.93\) (b) \(9.31\) (c) \(37.24\) (d) \(18.62\)
View solution Problem 14
Cadmium rods are used for which purpose? (a) to emit electrons (b) to adsorb neutrons (c) to emit neutrons (d) to absorb electrons
View solution