Question

Two radioactive nuclides \(A\) and \(B\) have half-lives \(50 \mathrm{~min}\) and \(10 \mathrm{~min}\) respectively. A fresh sample contains the nuclide of \(B\) to be eight time that of \(A\). How much time should elapse so that the number of nuclides of \(A\) becomes double of \(B\) ? (a) 30 (b) 40 (c) 50 (d) 100

Short answer

100 minutes

Step-by-step solution

Understand the half-life concept

Half-life is the time required for a quantity to reduce to half its initial value. The formula to calculate the remaining quantity of a radioactive substance after time t is given by the equation: \( N_t = N_0 (1/2)^{\frac{t}{T}} \), where \( N_t \) is the remaining quantity, \( N_0 \) is the initial quantity, \( t \) is the time elapsed, and \( T \) is the half-life of the substance.

Set up the initial scenario

Initially, the quantity of nuclide B is eight times that of A, which can be represented as: \( N_{B_0} = 8N_{A_0} \).

Define the condition for the desired scenario

We need to find the time when the quantity of nuclide A becomes double that of B. This is represented by the equation: \( N_{A_t} = 2N_{B_t} \).

Apply the half-life formula to each nuclide

Using the half-life formula for each nuclide with the initial condition from step 2, we have: \( N_{A_t} = N_{A_0} (1/2)^{\frac{t}{50}} \) and \( N_{B_t} = 8N_{A_0} (1/2)^{\frac{t}{10}} \).

Set up the equation from the defined condition

From the condition in step 3, substitute the expressions from step 4 into the equation: \( N_{A_0} (1/2)^{\frac{t}{50}} = 2[8N_{A_0} (1/2)^{\frac{t}{10}}] \). Simplify this equation to find t.

Simplify the equation and solve for t

Cancel out \( N_{A_0} \) from the equation and simplify to get \( (1/2)^{\frac{t}{50}} = 16 (1/2)^{\frac{t}{10}} \). Taking log base 1/2 of both sides results in \( \frac{t}{50} = 4 + \frac{t}{10} \). Solve this linear equation to find the value of t.

Calculate the elapsed time t

When you solve \( \frac{t}{50} = 4 + \frac{t}{10} \), you will get the linear equation in the form of \( t(\frac{1}{50} - \frac{1}{10}) = -4 \), which simplifies to \( -\frac{t}{25} = -4 \). Solving for t gives you t = 100 minutes.

Key concepts

Radioactive Decay

Radioactive decay is a spontaneous process by which an unstable atomic nucleus loses energy by emitting radiation. This remarkable phenomenon is integral to understanding nuclear chemistry and is foundational in the study of physical chemistry for competitive exams like JEE. In the context of half-life problems, it's essential to realize that during radioactive decay, the amount of a radioactive isotope decreases over time in a predictable manner.

The decay follows an exponential function, where the time required for half of the original quantity of radioactive substance to decay is called its 'half-life'. This characteristic allows scientists and students alike to calculate the remaining amount of a substance after a given period.

Understanding the mechanics of radioactive decay is invaluable not only for answering specific JEE questions but also for comprehending broader themes in nuclear physics and the natural radioactive processes occurring in our environment.

Nuclear Chemistry

Nuclear chemistry deals with the reactions and changes that occur in the nucleus of atoms. It encompasses a broad range of topics, including radioactivity, fission, and fusion processes. Within the JEE physical chemistry syllabus, nuclear chemistry is a pivotal unit because it explains both the energy produced in stars and the principles behind nuclear reactors.

When tackling half-life problems, one must understand that the stability of a nuclide, the particular type or species of nucleus characterized by its number of protons and neutrons, dictates its potential to undergo decay. These problems require a grasp of the delicate balance within the nucleus and the concept of binding energy which is the energy required to split a nucleus into its individual protons and neutrons.

By mastering nuclear chemistry, students gain insights into the power and energy harnessed from the smallest particles in the universe, setting a strong foundation for further studies or careers in chemistry, physics, and engineering.

Exponential Decay

Exponential decay is a specific mathematical model that describes the process of reducing an amount by a consistent percentage rate over time. This concept is closely linked to half-life problems, as radioactive substances decrease in quantity following this exact model. In a typical exponential decay equation, the value decreases at a rate proportional to its current value.

Mathematically, the formula to represent this decrease is often written as:
\[ N_t = N_0 \times (1/2)^{\frac{t}{T}} \]
where:

• \( N_t \) is the quantity at time \( t \),
• \( N_0 \) is the initial quantity,
• \( T \) is the half-life, and
• \( t \) is the elapsed time.

This model is not only applicable in nuclear chemistry but also in fields like finance, biology, and geology. For JEE aspirants, understanding exponential decay is crucial as it routinely features in both conceptual and numerical problem-solving.

Physical Chemistry for JEE

Physical chemistry is a significant part of the JEE syllabus. It merges the principles of physics and chemistry to explain how matter behaves on a molecular and atomic level. Concepts such as radioactive decay and exponential decay are central to this branch of chemistry, especially when discussing the kinetics of reactions and the energetic processes in atomic nuclei.

For JEE applicants, gaining a thorough knowledge of physical chemistry is mandatory as it's replete with both conceptual understanding and quantitative calculations. Half-life problems, such as the one explored in our exercise example, are a staple in JEE exams and demand a comprehensive understanding of the underlying principles to be solved effectively.

Thus, students need to focus on developing a robust conceptual framework along with adeptness in problem-solving techniques by regularly practicing and applying those concepts to exercises, which prepares them for the diverse challenges of the JEE exams.