Question

\({ }_{90} \mathrm{Th}^{234}\) disintegrate to give \({ }_{82} \mathrm{~Pb}^{206}\) as the final product. Total no. of \(\alpha\) and \(\beta\) particles emitted out during this process are : (a) 6 (b) 7 (c) 8 (d) 13

Short answer

A total of 13 particles, with 7 alpha particles and 6 beta particles, are emitted.

Step-by-step solution

Identify the Starting and Final Nuclei

First, identify the starting and final nuclei in the nuclear reaction. The initial nucleus is Thorium-234 ({ }_{90} Th^{234}), and the final nucleus is Lead-206 ({ }_{82} Pb^{206}).

Determine Change in Atomic Number

Calculate the change in atomic number (Z) from the parent nucleus to the final nucleus. The change in atomic number is 90 - 82 = 8. This indicates that 8 alpha particles must have been emitted since each alpha particle emission decreases the atomic number by 2.

Determine Change in Mass Number

Calculate the decrease in mass number (A) from Th^{234} to Pb^{206}. The change in mass number is 234 - 206 = 28. Since each alpha particle decreases the mass number by 4, dividing the total change in mass number by 4 gives the number of alpha particles emitted: 28/4 = 7 alpha particles.

Calculate Number of Beta Particles Emitted

We know beta decay doesn't change the mass number but increases the atomic number by 1. We have already accounted for the decrease in atomic number due to alpha decay (7 times), so the additional atomic number decrease has to come from the emission of beta particles. To get from the atomic number after alpha decay (90 - 7*2 = 76) down to 82, we need to emit 6 beta particles: 76 + 6 = 82.

Add Alpha and Beta Emissions

Sum the number of alpha and beta particles emitted. The total number of alpha particles is 7 and beta particles is 6, so the total emission is 7 + 6 = 13.

Key concepts

Alpha Particle Emission

Alpha particle emission is a common mode of radioactive decay in which an unstable nucleus emits an alpha particle. An alpha particle is essentially a helium nucleus, consisting of two protons and two neutrons, symbolically represented as (_2^4He). When an alpha particle is emitted from a nucleus, the parent atom’s mass number decreases by four (since it loses two protons and two neutrons) and its atomic number decreases by two (because it loses two protons), leading to the formation of a new element.
In our exercise, the transition from Thorium-234 to Lead-206 requires several alpha emissions. Historically, such reactions are significant not only in academic exercises but also for understanding natural radioactive decay series in physical chemistry, which is an essential topic in the Joint Entrance Examination (JEE) syllabus for advanced studies in physical chemistry.
By following the decay series, we can track the stability of atoms and the emergence of new elements, highlighting the practical aspects of alpha decay in nuclear chemistry.

Beta Particle Emission

In contrast to alpha decay, beta particle emission involves the transformation of a neutron into a proton within an atomic nucleus, leading to the emission of an electron, known as a beta particle. This process increases the atomic number by one while leaving the mass number unchanged.
In terms of particle physics, beta decay is actually the emission of an electron (_(-1)^0e) or a positron (_1^0e), accompanied by a neutrino or an antineutrino, respectively. This intricate process showcases one of the many ways an unstable nucleus can attain stability. In our problem's context, the presence of beta particles hinted that there were additional modifications to Thorium-234, as beta emission is necessary to adjust the atomic number after alpha emissions have altered both the atomic and mass numbers.
Understanding beta decay helps students tackling physical chemistry JEE problems to appreciate the interplay of nuclear forces that govern atomic stability and the role of weak nuclear forces in particle interactions.

Nuclear Reaction Calculations

Solving nuclear decay problems often involves calculating changes in atomic and mass numbers. While working through such problems, it's essential to establish a systematic approach, starting with identifying the parent and daughter isotopes. After pinpointing these isotopes, students should calculate the change in the atomic number to determine the number of alpha and beta particles emitted.
In the given exercise, the correct approach involved calculating the difference in atomic numbers, which indicated the number of alpha emissions, and then calculating the resultant change this would have caused in the atomic number to find the requisite number of beta emissions. Such calculations form the bedrock of nuclear chemistry and are especially vital for exams like the JEE, where understanding underlying concepts translates to solving complex multiple-step problems efficiently.

Physical Chemistry JEE

Physical chemistry for JEE encompasses a broad spectrum of topics, including nuclear chemistry, which plays a pivotal role in understanding radioactive decay series, nuclear reactors and applications of radioactivity. The Joint Entrance Examination (JEE) assesses a student's proficiency and understanding of core science principles, which are fundamental in advanced scientific education and research.
Alpha and beta particle emissions are not just isolated phenomena but integrated topics within physical chemistry. For JEE aspirants, mastering such topics is crucial for tackling a variety of theoretical and calculation-based questions that evaluate their analytical skills and problem-solving capabilities. The exercise provided is an example of how real-world nuclear reactions are distilled into textbook problems, helping students navigate the complexities of physical chemistry in a structured and relatable manner.