Problem 89 - Master Resource Book in JEE Main Physics

Question

In the uranium radioactive series, the initial nucleus is \({ }_{92} \mathrm{U}^{238}\) and that the final nucleus is \({ }_{82} \mathrm{~Pb}^{206}\). When uranium nucleus decays to lead, the number of \(\alpha\)-particle and \(\beta\)-particles emitted are (a) \(8 \alpha, 6 \beta\) (b) \(6 \alpha, 7 \beta\) (c) \(6 \alpha, 8 \beta\) (d) \(4 \alpha, 3 \beta\)

Step-by-Step Solution

Answer

(a) 8 \(\alpha\), 6 \(\beta\) particles are emitted.

Step 1: Understand Alpha Decay

In an alpha decay process, an \( \alpha \)-particle, which can be represented as \( _2^4\text{He} \), is emitted. This decreases the atomic number by 2 and the mass number by 4. That is, if an atom \( _Z^A\text{X} \) undergoes \( \alpha \)-decay, it transforms into \( _{Z-2}^{A-4}\text{X'} \).

Step 2: Understand Beta Decay

In beta decay, a \( \beta \)-particle, which is an electron (or a positron depending on the decay) is emitted. Typically, a neutron is converted into a proton, increasing the atomic number by 1 without changing the atomic mass. That is, \( _Z^A\text{X} \) becomes \( _{Z+1}^A\text{X}' \).

Step 3: Set Up the Problem with Uranium and Lead

The initial nucleus is \( _{92}\text{U}^{238} \) and the final nucleus is \( _{82}\text{Pb}^{206} \). We need to find how many \( \alpha \)-particles and \( \beta \)-particles are emitted in this transformation.

Step 4: Calculate the Number of Alpha Particles

Since each \( \alpha \)-particle decreases the atomic number by 2, to transform \( _{92}\text{U} \) to \( _{82}\text{Pb} \), you must emit \( x \) \( \alpha \)-particles. Set up the equation for the atomic mass: \( 238 - 4x = 206 \).Solving: \[ 238 - 4x = 206 \]\[ 4x = 238 - 206 \]\[ 4x = 32 \]\[ x = \frac{32}{4} = 8 \]So, 8 \( \alpha \)-particles are emitted.

Step 5: Calculate the Number of Beta Particles

Each \( \alpha \)-decay reduces the atomic number by 2, thus for 8 \( \alpha \)-decay, the atomic number reduces by \( 2 \times 8 = 16 \). Therefore, from 92, the atomic number would drop to 76 if only \( \alpha \)-decay happened. Since the final atomic number must be 82, an increase of \( 82 - 76 = 6 \) is needed due to \( \beta \)-decay.Thus, 6 \( \beta \)-particles are emitted.

Step 6: Conclusion

Therefore, in the full decay sequence from \( _{92} \text{U}^{238} \) to \( _{82} \text{Pb}^{206} \), there are 8 \( \alpha \)-particles and 6 \( \beta \)-particles emitted, matching option (a).

Key Concepts

Alpha Decay

In the realm of nuclear decay, alpha decay involves the emission of an alpha particle, which is essentially a helium nucleus, composed of two protons and two neutrons. When an unstable atom undergoes alpha decay, it sheds an alpha particle, resulting in the creation of a new element. This process causes a decrease in both the atomic number by 2 and the mass number by 4.
For example, if a nucleus of the form \(_Z^A\text{X}\) undergoes alpha decay, it changes into \(_{Z-2}^{A-4}\text{Y}\). The reason behind alpha decay is the quest for greater stability. The nucleus releases an alpha particle to lower the overall energy and achieve a more stable state.

Beta Decay

Beta decay occurs when a neutron in a radioactive nucleus is transformed into a proton, which results in the emission of a beta particle. This beta particle can be an electron or a positron, depending on the type of beta decay. In the case of beta-minus decay, an electron and an antineutrino are emitted, while in beta-plus decay, a positron and a neutrino are released.
During beta decay, the atomic number of the element increases by 1, but the atomic mass remains unchanged. This is because only a proton-neutron transformation occurs. Therefore, a nucleus \(_Z^A\text{X}\) after beta decay changes to \(_{Z+1}^A\text{Y}\).
Beta decay is critical for reaching a stable configuration in the nucleus after other decay processes, like alpha decay, have occurred. It helps to manage the proton-to-neutron ratio, which is key to nuclear stability.

Uranium-Lead Decay Series

The uranium-lead decay series is a chain reaction of radioactive decay processes that starts with uranium-238 and ends with lead-206. This decay sequence is significant in geochronology for dating rocks and understanding Earth's history.
In this series, uranium-238 slowly breaks down over billions of years to eventually become stable lead-206. It involves a series of alpha and beta decays. Specifically, it emits eight alpha particles and six beta particles as observed in the conversion from uranium \(_{92}^{}U^{238}\) to lead \(_{82}^{}Pb^{206}\).

• Alpha decays contribute to a reduction in atomic number by 16 (2 per decay)
• Beta decays help in regaining 6 atomic numbers

This decay series aids our understanding of nuclear reactions and the concept of half-life, a critical aspect of radioactivity.

Nuclear Reactions

Nuclear reactions involve a transformation in the composition, energy, or structure of atomic nuclei. Unlike chemical reactions, which involve the sharing or transfer of electrons, nuclear reactions occur at the core of the atom. They are responsible for releases or absorptions of significant amounts of energy.
A typical nuclear reaction can be represented as: \(_{A_1}^{Z_1} ext{X} + _{A_2}^{Z_2} ext{Y} \rightarrow _{A_3}^{Z_3} ext{W} + E\), where \(E\) represents the energy released or absorbed. Nuclear decay processes such as alpha and beta decays are examples of spontaneous nuclear reactions, wherein unstable nuclei emit particles to shift to a more stable state.
Other notable nuclear reactions include fission, where a heavy nucleus splits into smaller nuclei, and fusion, where light nuclei unite to form a heavier nucleus. Understanding these processes provides insight into both natural phenomena and technological applications, like nuclear power generation and medical imaging.