Chapter 25: Problem 10
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) \(2 \sqrt{2}\)
Short Answer
Expert verified
The ratio of their de-Broglie wavelengths is \( 2\sqrt{2} \) (Option d).
Step by step solution
01
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.
02
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, \( q_\alpha = 2e \).
03
Calculating de-Broglie Wavelength for Proton
For a proton, the momentum is \( p_p = \sqrt{2m_p eV} \). Therefore, the de-Broglie wavelength is \( \lambda_p = \frac{h}{\sqrt{2m_p eV}} \).
04
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 \( \lambda_\alpha = \frac{h}{\sqrt{4m_\alpha eV}} \).
05
Finding the Ratio of Wavelengths
The ratio \( \frac{\lambda_p}{\lambda_\alpha} \) is given by:\[ \frac{\lambda_p}{\lambda_\alpha} = \frac{\frac{h}{\sqrt{2m_p eV}}}{\frac{h}{\sqrt{4m_\alpha eV}}} = \frac{\sqrt{4m_\alpha eV}}{\sqrt{2m_p eV}} = \sqrt{\frac{2m_\alpha}{m_p}} \].
06
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:\[ \frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{2 \times 4m_p}{m_p}} = \sqrt{8} = 2\sqrt{2} \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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.
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.
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.