Question

The isotope \(\frac{234}{92} \mathrm{U}\) is known to undergo a series of five successive alpha-particle emissions before ending up as the isotope \(\frac{214}{82} \mathrm{Pb}\) (at which point the isotope undergoes a series of beta-particle emissions before finally reaching stability). Write nuclear equations for the series of alpha emissions undergone by \(\frac{244}{92} \mathrm{U}\) as it decays to \(_{82}^{214} \mathrm{Pb}\)

Short answer

The nuclear equations for the series of five alpha emissions undergone by \(\frac{234}{92} \mathrm{U}\) are: 1. \(\frac{234}{92} \mathrm{U} \rightarrow \frac{4}{2} \mathrm{He} + \frac{230}{90} X\) 2. \(\frac{230}{90} X\rightarrow \frac{4}{2} \mathrm{He} + \frac{226}{88} X\) 3. \(\frac{226}{88} X \rightarrow \frac{4}{2} \mathrm{He} + \frac{222}{86} X\) 4. \(\frac{222}{86} X \rightarrow \frac{4}{2} \mathrm{He} + \frac{218}{84} X\) 5. \(\frac{218}{84} X \rightarrow \frac{4}{2} \mathrm{He} + \frac{214}{82} \mathrm{Pb}\)

Step-by-step solution

Write down the initial isotope

The initial isotope given is \(\frac{234}{92} \mathrm{U}\).

Write the alpha particle symbol

An alpha particle is represented by the symbol \(\frac{4}{2} \mathrm{He}\).

Write nuclear equations for each alpha emission

As there are five successive alpha emissions, we will write the nuclear equation for each step. 1. First alpha emission: \(\frac{234}{92} \mathrm{U} \rightarrow \frac{4}{2} \mathrm{He} + X_{1}\) Determine \(X_{1}\): Mass number conservation: \(234 = 4 + A_{1}\) Atomic number conservation: \(92 = 2 + Z_{1}\) Solving the conservation equations, we get: \(A_{1} = 230\), \(Z_{1} = 90\) Thus, \(X_{1} = \frac{230}{90}\) 2. Second alpha emission: \(\frac{230}{90} \rightarrow \frac{4}{2} \mathrm{He} + X_{2}\) Determine \(X_{2}\): Mass number conservation: \(230 = 4 + A_{2}\) Atomic number conservation: \(90 = 2 + Z_{2}\) Solving the conservation equations, we get: \(A_{2} = 226\) \(Z_{2} = 88\) Thus, \(X_{2} = \frac{226}{88}\) 3. Third alpha emission: \(\frac{226}{88} \rightarrow \frac{4}{2} \mathrm{He} + X_{3}\) Determine \(X_{3}\): Mass number conservation: \(226 = 4 + A_{3}\) Atomic number conservation: \(88 = 2 + Z_{3}\) Solving the conservation equations, we get: \(A_{3} = 222\) \(Z_{3} = 86\) Thus, \(X_{3} = \frac{222}{86}\) 4. Fourth alpha emission: \(\frac{222}{86} \rightarrow \frac{4}{2} \mathrm{He} + X_{4}\) Determine \(X_{4}\): Mass number conservation: \(222 = 4 + A_{4}\) Atomic number conservation: \(86 = 2 + Z_{4}\) Solving the conservation equations, we get: \(A_{4} = 218\) \(Z_{4} = 84\) Thus, \(X_{4} = \frac{218}{84}\) 5. Fifth alpha emission: \(\frac{218}{84} \rightarrow \frac{4}{2} \mathrm{He} + X_{5}\) Determine \(X_{5}\): Mass number conservation: \(218 = 4 + A_{5}\) Atomic number conservation: \(84 = 2 + Z_{5}\) Solving the conservation equations, we get: \(A_{5} = 214\) \(Z_{5} = 82\) Thus, \(X_{5} = \frac{214}{82} \mathrm{Pb}\) Following five alpha emissions, the initial isotope \(\frac{234}{92} \mathrm{U}\) has decayed to the isotope \(\frac{214}{82} \mathrm{Pb}\).

Key concepts

Alpha-Particle Emission

Alpha-particle emission is a type of nuclear decay where an unstable nucleus releases an alpha particle to become more stable. An alpha particle is essentially a helium nucleus, represented by the symbol \(\frac{4}{2} \mathrm{He}\). It consists of two protons and two neutrons, making it a positively charged particle with significant mass compared to other particles like beta particles. During nuclear decay, releasing an alpha particle reduces the mass number of the original isotope by 4, and its atomic number by 2.
Alpha particle emission is commonly encountered in heavy elements like uranium and thorium. These emissions occur as the nucleus seeks a more stable configuration, typically leading to a series of decay events known as a decay chain. For example, when \(\frac{234}{92} \mathrm{U}\) undergoes decay through successive alpha emissions, it eventually becomes \(\frac{214}{82} \mathrm{Pb}\).
Alpha decay plays a crucial role in the natural conversion of unstable elements into stable ones. Understanding this concept helps us comprehend how elements change over time in natural processes.

Nuclear Equations

Nuclear equations allow us to describe the changes occurring in an atomic nucleus as a result of nuclear decay. These equations help in following the conservation laws of mass and charge throughout the process. A typical nuclear equation for alpha decay shows the parent nucleus on the left and the resulting daughter nucleus and emitted alpha particle on the right.
In the case of uranium \(\frac{234}{92} \mathrm{U}\) undergoing alpha decay, the equation would be:

• \(\frac{234}{92} \mathrm{U} \rightarrow \frac{4}{2} \mathrm{He} + \frac{230}{90} X\)

Here, the parent nucleus loses 4 from its mass number and 2 from its atomic number with each alpha emission. The mathematical beauty of nuclear equations lies in their consistent adherence to the laws of conservation, ensuring that calculations of mass numbers and atomic numbers (protons) remain balanced on both sides of the equation.
By writing nuclear equations step by step for a series of alpha decays, one can track the transformation of elements over multiple decay events, fully understanding every stage of the nuclear transformation process.

Uranium Isotopes

Uranium isotopes are an interesting subject in nuclear chemistry. Uranium is a heavy element, known for containing isotopes with different numbers of neutrons, most commonly \(\frac{235}{92} \mathrm{U}\) and \(\frac{238}{92} \mathrm{U}\). However, \(\frac{234}{92} \mathrm{U}\) is also a significant isotope considered in decay studies. Each isotope of uranium has unique properties, and their behavior during nuclear reactions can be notably different.
Isotopes are identified by their mass numbers, which is the sum of protons and neutrons in the nucleus. Despite having the same number of protons (92 in uranium), these isotopes differ due to their neutron count. This difference affects their stability and the way they decay. For instance, \(\frac{234}{92} \mathrm{U}\) undergoes a series of alpha decays, making it a crucial part of discussing decay chains in nuclear reactions.
Understanding uranium isotopes and their decay helps us in multiple domains, such as nuclear energy production and radioactive dating techniques. With 92 protons, these isotopes are among the heaviest naturally occurring elements and provide insight into processes that shape the composition of our planet over geological timescales.