In a radioactive decay chain, \({ }_{90}^{232} \mathrm{Th}\) nucleus decays to \({ }_{82}^{212} \mathrm{~Pb}\) nucleus. Let \(N_{\alpha}\) and \(N_{\beta}\) be the number of \(\alpha\) and \(\beta^{-}\) particles, respectively, emitted in this decay process. Which of the following statements is (are) true? (A) \(N_{\alpha}=5\) (B) \(N_{\alpha}=6\) (C) \(N_{\beta}=2\) (D) \(N_{\beta}=4\)

Alpha decay is a type of radioactive decay wherein an unstable nucleus emits an alpha particle, essentially a helium-4 nucleus, to become a more stable element. This process results in the reduction of the mass number (A) of the parent nuclide by four units and the atomic number (Z) by two units.

For example, when \( { }_{90}^{232} \mathrm{Th} \) undergoes alpha decay, it ejects an alpha particle (\( { }_{2}^{4} \mathrm{He} \) nucleus) and transforms into a new nucleus with a mass number of 228 and an atomic number of 88. The equation for this decay process is: \[ { }_{90}^{232} \mathrm{Th} \rightarrow { }_{88}^{228} \mathrm{Ra} + { }_{2}^{4} \mathrm{He} .\]

In the context of JEE Advanced Physics, understanding alpha decay is crucial, as it forms a part of the nuclear physics curriculum and is often tested in this prestigious examination.

Beta decay is another form of radioactive decay, in which a nucleus emits a beta particle, which can be either a high-speed electron or positron. During \(\beta^{-}\) decay, a neutron transforms into a proton, an electron, and an antineutrino. The emitted electron is called the beta particle. This decay results in an increase in the atomic number by one, while the mass number remains unchanged.

The decay of \( { }_{90}^{232} \mathrm{Th} \) involves both alpha and beta decays to eventually form \( { }_{82}^{212} \mathrm{~Pb} \). Since the mass number doesn't change during beta decay, it's the alpha decays that account for the mass number change from 232 to 212, but beta decays increase the atomic number from the intermediate product to lead's atomic number. This is key to solving problems like the one under consideration and is often tested in competitive exams like JEE Advanced, making it a vital concept for nuclear physics topics.

Nuclear physics is a branch of physics that deals with the constituents and interactions of atomic nuclei. Core topics include the study of radioactivity, fusion, fission, and the behavior of nuclei under various conditions. Principles of nuclear physics are applied in various fields such as medical imaging, power generation, and understanding cosmological phenomena.

Within the scope of JEE Advanced Physics, students must comprehend how fundamental interactions manifest in phenomena like radioactive decay chains. For instance, they need to solve problems based on the sequential transformation of nuclei involving alpha and beta decays, requiring a clear grasp of how mass numbers and atomic numbers change during these processes. As these topics are complex and require high analytical skills, they are perfect candidates for challenging questions in competitive exams like JEE Advanced.

The Joint Entrance Examination (JEE) Advanced Physics section assesses a student’s understanding of fundamental physics concepts at an advanced level. It is known for its challenging problems that test application, analytical thinking, and conceptual clarity. Radioactive decay chains, involving alpha and beta particles, regularly feature in the exam.

Answering these types of questions correctly requires the application of nuclear physics principles. As demonstrated in the original exercise, knowing the atomic number and mass number adjustments for each decay type is crucial for deducing the number of alpha and beta particles emitted during a decay sequence. Being adept in such problems facilitates success in JEE Advanced Physics and is critical for aspiring engineers and scientists who wish to enter premier institutes in India.