Solving Physical Chemistry Problems for JEE: A Comprehensive Guide to Alpha Particle Scenarios

Chemistry, particularly physical chemistry, often poses intriguing problems to students preparing for JEE (Joint Entrance Examination). Let's take a closer look at how to solve a question involving alpha particles through the laws of conservation of energy.

Understanding the Basics

In physical chemistry, the concept of energy conservation is crucial. This principle states that the total energy of a closed system remains constant over time. For this particular problem, we are dealing with a scenario where an alpha particle interacts with the nucleus of an atom.

Problem Setup and Initial Conditions

In the initial setup, we start with the alpha particle at infinity (a distance essentially considered to be infinite), where the potential energy is zero. The alpha particle is then given an initial velocity ( V_0 ).

Laws of Conservation of Energy

According to the law of conservation of energy, the total energy of the system (which includes both kinetic and potential energy) remains constant throughout the process. Initially, the kinetic energy of the alpha particle is given by:

$$text{KE} frac{1}{2}mv^2$$

where ( m ) is the mass of the alpha particle and ( v ) is its initial velocity ( V_0 ).

Key Points of Interaction

As the alpha particle approaches the nucleus, its potential energy increases, while its kinetic energy decreases because it is moving against a repulsive force. The point at which the alpha particle comes to a complete stop (i.e., its kinetic energy becomes zero) is the closest it can get to the nucleus. At this point, the total energy is entirely in the form of potential energy.

Let's denote the closest distance the alpha particle reaches to the nucleus as ( R ).

Calculating the Distance ( R )

The potential energy at a distance ( R ) from the nucleus can be calculated using:

$$text{PE} frac{kQq}{R}$$

where ( k ) is Coulomb's constant, ( Q ) is the charge of the nucleus, and ( q ) is the charge of the alpha particle. Since the initial kinetic energy is equal to the potential energy at the closest point:

$$frac{1}{2}mv_0^2 frac{kQq}{R}$$

Solving for ( R ) gives:

$$R frac{2kQq}{mv_0^2}$$

Generalization to Similar Problems

When dealing with similar problems involving alpha particles at different distances from the nucleus, the same principle applies. Let's say we want to find the closest distance ( R_1 ) when the velocity is not zero but some other value ( V_1 ).

Applying the Law of Conservation of Energy

In this scenario, the initial conditions are the same as before, but the final conditions include a non-zero velocity ( V_1 ). The potential energy at distance ( R_1 ) is given by:

$$text{PE} frac{kQq}{R_1}$$

Using the law of conservation of energy, the total initial energy (kinetic potential) equals the total final energy (kinetic potential) at distance ( R_1 ).

Equating and Solving

The total initial energy is:

$$text{Total Initial Energy} frac{1}{2}mv_0^2 0$$

The total final energy at distance ( R_1 ) is:

$$text{Total Final Energy} frac{1}{2}mv_1^2 frac{kQq}{R_1}$$

Equating the two gives:

$$frac{1}{2}mv_0^2 frac{1}{2}mv_1^2 frac{kQq}{R_1}$$

Solving for ( R_1 ) gives:

$$R_1 frac{kQq}{frac{1}{2}m(v_0^2 - v_1^2)}$$

Conclusion

The problem of an alpha particle interacting with a nucleus is a classic application of the law of conservation of energy. By understanding the relationship between kinetic and potential energy, we can solve such problems accurately and efficiently.

To wrap up, here are some key takeaways:

Initial conditions at infinity (zero potential energy). Kinetic energy is converted to potential energy as the particle approaches the nucleus. Applying the law of conservation of energy allows us to find the closest distance to the nucleus. The same principle applies to similar problems involving different velocities.

Keep practicing and understanding these fundamentals will help you tackle more complex problems in physical chemistry for JEE.