Problem 146 - A Complete Resource Book in Chemistry for JEE Main

Question

In the transformation of \({ }_{92} \mathrm{U}^{238}\) to \({ }_{92} \mathrm{U}^{234}\), if one emission is an \(\alpha\) particle, what should be the other emission(s)? \([2006]\) (a) two \(\beta\) (b) two \(\beta\) and one \(\beta^{+}\) (c) one \(\beta\) - and one \(\gamma\) (d) one \(\beta\) and one \(\beta\)

Step-by-Step Solution

Answer

The transformation involves the emission of one \( \alpha \) particle and two \( \beta \) particles.

Step 1: Understand the Problem

In the problem, we are given a uranium nuclear transformation from \( {}_{92} \mathrm{U}^{238} \) to \( {}_{92} \mathrm{U}^{234} \). We know an \( \alpha \) particle emission has occurred. We need to determine what other type of emission(s) complete this transformation.

Step 2: Understanding Alpha Emission

An \( \alpha \) particle is a helium nucleus consisting of 2 protons and 2 neutrons. Thus, when a nucleus emits an \( \alpha \) particle, its atomic number decreases by 2, and its mass number decreases by 4. So, from \( {}_{92} \mathrm{U}^{238} \), it becomes \( {}_{90} \text{Th}^{234} \).

Step 3: Determine the Need for Beta Emissions

After alpha emission, the resultant element is \( {}_{90} \text{Th}^{234} \). We need to get back to uranium with the same atomic number as the original uranium (92), but now with a mass number 234.\(\beta \) emission increases the atomic number by 1, without changing the mass number.

Step 4: Calculating Required Beta Emissions

Starting from \( {}_{90} \text{Th}^{234} \), we need to increase the atomic number twice to return to uranium (with atomic number 92). Therefore, two \( \beta \) emissions are required to convert \({}_{90} \text{Th}^{234}\) to \( {}_{92} \mathrm{U}^{234} \).

Step 5: Conclusion on Emissions

The transformation from \( {}_{92} \mathrm{U}^{238} \) to \( {}_{92} \mathrm{U}^{234} \) involves one \( \alpha \) emission followed by two \( \beta \) emissions. Thus, the correct answer is option (a): two \( \beta \) emissions.

Key Concepts

Alpha Decay

Alpha decay is a type of nuclear transformation where an atom emits an alpha particle, which consists of 2 protons and 2 neutrons. This emission causes the parent atom to lose 4 units from its mass number and 2 units from its atomic number, resulting in the formation of a new element. For example, when uranium-238 undergoes alpha decay, it emits an alpha particle and transforms into thorium-234. The new element has two fewer protons and a corresponding change in its chemical identity. This process is significant because it helps stabilize heavy nuclei by reducing the overall size of the nucleus. In terms of notation, the alpha particle is often represented as \( {}_2^4 ext{He} \) or simply \( \alpha \). It’s important to remember this transformation while working with nuclear equations, as it always involves this consistent change in atomic and mass numbers.

Beta Decay

Beta decay is another common type of nuclear decay. It occurs when a nucleus has an imbalance in its neutron-to-proton ratio. In beta decay, a neutron is converted to a proton, releasing a beta particle (an electron or positron) in the process. Unlike alpha decay, beta decay does not change the mass number, but it does increase the atomic number by one if it is a beta-minus \((\beta^-)\) decay. This is vital to understand when dealing with isotopes that undergo beta decay transformations.

• Beta-minus \( (\beta^-) \): Converts a neutron to a proton and emits an electron.
• Beta-plus \( (\beta^+) \): Converts a proton to a neutron and emits a positron.

In our given exercise, two beta-minus decays were necessary after the alpha decay to change thorium-234 into uranium-234, increasing the atomic number back up to 92 while keeping the mass number at 234. Beta decay is especially interesting because it showcases the concept of weak nuclear force, one of the four fundamental forces in physics.

Uranium Isotopes

Uranium isotopes are nuclei of the element uranium that have the same number of protons but different numbers of neutrons. Uranium is well-known for its use in nuclear energy and weapons, primarily due to its radioactive isotopes such as uranium-235 and uranium-238. Each isotope has distinct nuclear properties and stability.
- **Uranium-238**: The most abundant uranium isotope in nature, it is not directly used for energy generation but is important for breeding plutonium-239.
- **Uranium-235**: Much less common, it is highly fissionable and critical for maintaining nuclear chain reactions. Understanding the properties of uranium isotopes is crucial for fields like nuclear physics and engineering. These isotopes differ mainly in their decay pathways, such as alpha and beta decay, which can lead to the formation of different elements over time.

Chemical Reactions in Nuclear Physics

In nuclear physics, chemical reactions refer to the changes that occur within the nucleus of an atom. Unlike ordinary chemical reactions that involve the interaction of electrons and the formation of new substances, nuclear reactions involve alterations in the atomic nucleus and often lead to the transformation of one element into another. Key characteristics of nuclear reactions include:

• Involvement of subatomic particles like protons, neutrons, and electrons.
• Changes in an atom's nucleus can result in new elements forming or nuclear energy being released.
• Processes such as fission, fusion, and radioactive decay (like alpha and beta decay) are examples of nuclear reactions.

Nuclear reactions are fundamentally different from chemical reactions due to the energies involved. While chemical reactions are concerned with electron exchanges, nuclear reactions harness and change the power within the nucleus, explaining their significant impact in areas like energy production and medical applications.