\begin{aligned} &\text { How many } \alpha-\text { and } \beta \text { -particles will be emitted when }{ }_{90} \mathrm{Th}^{232} \text { changes }\\\ &\text { into }{ }_{82} \mathrm{~Pb}^{208} \text { ? } \end{aligned}

The concept of alpha particle emission is pivotal in understanding many nuclear reactions, including radioactive decay processes. An alpha particle is essentially a helium nucleus composed of two protons and two neutrons, which makes it a hefty particle with a substantial impact on the atomic mass and atomic number of the decaying nucleus.

When an unstable nucleus emits an alpha particle, there is a significant change in its composition: the atomic mass decreases by four units since the emitted alpha particle carries away two protons and two neutrons, and the atomic number decreases by two because it is defined by the number of protons in the nucleus. These changes transform the original element into a new one.

In our exercise example, each emission of an alpha particle by Thorium-232 results in a stepwise approach towards becoming the more stable Lead-208. It's a balancing act between mass and charge that leads to the eventual stabilization of the nucleus.

Beta particle emission is another form of radioactive decay, where the emitted particle is an electron in the case of beta-minus decay, or a positron in beta-plus decay. Specifically, in beta-minus decay, a neutron is transformed into a proton with the accompanying release of an electron, which is the beta particle, and an antineutrino. Conversely, in beta-plus decay, a proton is converted into a neutron with the emission of a positron and a neutrino.

The key characteristic of beta decay is that it alters the atomic number of the nucleus—the count of protons—without affecting the atomic mass, since the mass of an electron or positron is negligible compared to the nucleus. This means that beta emission can change one chemical element into another by changing the number of protons while maintaining the same mass number.

Incorporating this into our exercise, it's clear that beta emissions are necessary to account for changes in atomic number without the reduction of atomic mass, as occurs in the latter stages of the Thorium to Lead transformation.

Nuclear reaction analysis is a sophisticated method used to ascertain the sequence of events in a nuclear transformation. By meticulously calculating the changes in atomic mass and atomic number during decay processes, scientists can deduce the types and quantities of particles emitted. This deduction follows strict rules of conservation; both the total mass and charge must be conserved in the process.

To approach a nuclear reaction analytically, you must first become knowledgeable about different types of decay and understand the significance of the atomic mass and atomic number in describing the characteristics of an element. The analysis often involves a step-by-step sequence of logical calculations, highlighting the importance of accurate bookkeeping in terms of mass and proton count.

In educational settings, such as the IIT-JEE problem solving, this analysis is critical. It necessitates a comprehensive understanding and careful calculation, especially when numbers seem not to tally. The ability to identify and rectify discrepancies, such as in our exercise example, is a quintessential skill developed through practice.

Preparing for competitive exams like the Indian Institutes of Technology Joint Entrance Examination (IIT-JEE) involves tackling complex chemistry problems, including those on nuclear chemistry and radioactive decay. Highly challenging, these problems require a deep comprehension of underlying principles, the ability to apply concepts to novel situations, and high accuracy in numerical calculations.

Students are expected to navigate through intricate problems such as the calculation of alpha and beta particle emissions during nuclear decay, which not only test their understanding of the subject matter but also their problem-solving skills and attention to detail. Missteps in calculation can undermine the entire solution, stressing the importance of each step in the process.

Illustrating this, the textbook solution we’ve expounded upon serves as an essence of these complexities. It demonstrates the calibration required between theory and application, the precision demanded in numerical analysis, and the systematic troubleshooting needed when discrepancies arise—skills that are paramount for success in the IIT-JEE Chemistry examination.