Problem 10 - Master Resource Book in JEE Main Physics

Question

A proton and an \(\alpha\)-particle are accelerated through the same potential difference. The ratio of their de-Broglie wavelengths, \(\left(\frac{\lambda_{p}}{\lambda_{\alpha}}\right)\) is (a) \(\frac{1}{2 \sqrt{2}}\) (b) \(\underline{1}\) (c) 2 (d) \(2 \sqrt{2}\)

Step-by-Step Solution

Answer

The ratio of their de-Broglie wavelengths is \( 2\sqrt{2} \) (Option d).

Step 1: Understanding the de-Broglie Wavelength Formula

The de-Broglie wavelength \( \lambda \) of any particle is given by the equation \( \lambda = \frac{h}{p} \), where \( h \) is Planck's constant and \( p \) is the momentum of the particle. The momentum \( p \) is obtained from \( p = \sqrt{2mK} \) where \( m \) is the particle's mass and \( K \) is its kinetic energy.

Step 2: Relating Kinetic Energy to Potential Difference

When a charged particle is accelerated by a potential difference \( V \), it gains kinetic energy \( K = qV \), where \( q \) is the charge of the particle. For a proton, \( q_p = e \) (the charge of an electron), and for an \( \alpha \)-particle, \( q_\alpha = 2e \).

Step 3: Calculating de-Broglie Wavelength for Proton

For a proton, the momentum is \( p_p = \sqrt{2m_p eV} \). Therefore, the de-Broglie wavelength is \( \lambda_p = \frac{h}{\sqrt{2m_p eV}} \).

Step 4: Calculating de-Broglie Wavelength for \(\alpha\)-Particle

For an \( \alpha \)-particle, the momentum is \( p_\alpha = \sqrt{2m_\alpha (2e)V} = \sqrt{4m_\alpha eV} \). Therefore, the de-Broglie wavelength is \( \lambda_\alpha = \frac{h}{\sqrt{4m_\alpha eV}} \).

Step 5: Finding the Ratio of Wavelengths

The ratio \( \frac{\lambda_p}{\lambda_\alpha} \) is given by:\[ \frac{\lambda_p}{\lambda_\alpha} = \frac{\frac{h}{\sqrt{2m_p eV}}}{\frac{h}{\sqrt{4m_\alpha eV}}} = \frac{\sqrt{4m_\alpha eV}}{\sqrt{2m_p eV}} = \sqrt{\frac{2m_\alpha}{m_p}} \].

Step 6: Substituting Mass and Simplifying

The mass of an \( \alpha \)-particle is approximately \( 4 \) times the mass of a proton (\( m_\alpha \approx 4m_p \)). Substituting, we get:\[ \frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{2 \times 4m_p}{m_p}} = \sqrt{8} = 2\sqrt{2} \].

Key Concepts

Potential Difference

The potential difference plays a crucial role in the motion of charged particles, like protons and alpha particles. When a charged particle is subjected to a potential difference, it experiences a change in electric potential energy. This change can accelerate the particle. The amount of energy gained by the particle is known as its kinetic energy, which can be expressed as:

• For any particle with charge \( q \), the kinetic energy \( K \) gained is \( K = qV \), where \( V \) is the potential difference.

In our scenario, a proton with charge \( e \), and an alpha particle with charge \( 2e \), are both accelerated by the same potential difference. Despite being subjected to the same potential difference, different particles acquire different amounts of kinetic energy due to their unique charges. Understanding this concept helps us realize how kinetic energy and momentum relate to an electric potential which paves the way to uncovering deeper insights into particle physics.

Alpha Particle

An alpha particle, often symbolized as \( \alpha \), is a type of atomic particle found commonly in nuclear physics. It consists of two protons and two neutrons, making it a helium nucleus. Here are some key characteristics of alpha particles that are relevant:

• Charge: Being composed of two protons, an alpha particle has a total charge of \( +2e \), double that of a single proton.
• Mass: The mass of an alpha particle is roughly four times that of a single proton.
• Role in de-Broglie Wavelength: When calculating the de-Broglie wavelength of an alpha particle, its larger mass and double charge compared to a proton must be accounted for. This affects its momentum and ultimately its wavelength.

Recognizing these properties is essential when comparing the behaviors of alpha particles and protons, especially under the influence of external forces, such as an electric field caused by a potential difference.

Kinetic Energy

Kinetic energy is the energy a particle has due to its motion. For charged particles accelerated by a potential difference, kinetic energy can be calculated directly from the charge and potential difference.

• Formula: The kinetic energy \( K \) gained by a charged particle is given by \( K = qV \), where \( q \) is the charge and \( V \) is the potential difference.
• Impact on de-Broglie Wavelength: As kinetic energy increases, a particle's momentum increases, and this reduces its de-Broglie wavelength, following the relationship \( \lambda = \frac{h}{p} \), where \( p \) is momentum.
• Comparison: When a proton and an alpha particle are accelerated through the same potential difference, the alpha particle gains more kinetic energy due to its greater charge \( 2e \), potentially resulting in a different wavelength compared to the proton.

This concept of kinetic energy is pivotal in linking electric fields to physical movement, allowing us to delve into the quantum behaviors of particles.

Momentum Formula

Momentum is a key ingredient in determining the de-Broglie wavelength of a particle. The momentum \( p \) of a particle is defined by its mass and velocity, and it helps in calculating the wavelength through the relation \( \lambda = \frac{h}{p} \), where \( h \) is Planck's constant.

• Relation to Kinetic Energy: For a moving particle, \( p = \sqrt{2mK} \), where \( m \) is the mass and \( K \) the kinetic energy.
• Calculating for Proton and Alpha Particle: In the exercise, the momentum for a proton and an alpha particle are given by their respective masses and the energy gained from the applied potential difference.
• Impact on Wavelength: Higher momentum results in shorter wavelengths, illustrating the wave-like property of particles, crucial to quantum mechanics.

Understanding the momentum formula provides a bridge between classical mechanics and quantum mechanics, manifesting in the particle-wave duality through de-Broglie's hypothesis.