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A beam of protons and a beam of alpha particles (of mass \(6.64 \times 10^{-27} \mathrm{kg}\) and charge \(+2 e )\) are accelerated from rest through the same potential difference and pass through identical circular holes in a very thin, opaque film. When viewed far from the hole, the diffracted proton beam forms its first dark ring at \(15^{\circ}\) with respect to its original direction. When viewed similarly, at what angle will the alpha particle form its first dark ring?

Short Answer

Expert verified
For alpha particles, the first dark ring forms at approximately \( 10.61^{\circ} \).

Step by step solution

01

Understanding the Problem

To solve this problem, we need to find the angle at which the first dark ring is formed for alpha particles, using the given information about protons. The key concept here is that both protons and alpha particles undergo diffraction, and since they are accelerated by the same potential difference, they have the same kinetic energy after acceleration.
02

Applying Energy Conservation

Both the protons and alpha particles are accelerated from rest through the same potential difference, so they gain the same kinetic energy. This can be expressed as: \[ KE = qV, \] where \( q \) is the charge and \( V \) is the potential difference. Since \( q \) is \(+e\) for protons and \(+2e\) for alpha particles, their kinetic energies are proportional to their charges.
03

Calculating the Wavelengths Using de Broglie's Formula

For diffraction, we consider the de Broglie wavelength \( \lambda \), which is given by \[ \lambda = \frac{h}{\sqrt{2mKE}}, \] where \( h \) is Planck's constant, \( m \) is the mass, and \( KE \) is the kinetic energy. Since both types of particles have different masses and charges, but the same kinetic energy potential, we need to substitute these into the equation to find their respective wavelengths.
04

Relating Wavelength to First Dark Ring Angle

The angle \( \theta \) of the first dark ring is determined using the formula for diffraction: \[ \sin(\theta) = \frac{1.22 \lambda}{D}, \] where \( D \) is the diameter of the diffraction hole. We know the angle for protons, and we need to find it for alpha particles by relating their wavelengths.
05

Using the Angle and Mass Relationship

The ratio of the wavelength \( \lambda \) for alpha particles to protons can be given by the ratio of their masses, adjusted for their respective chargers. Given \( \sin(\theta_{\alpha})/\sin(\theta_{p}) = \lambda_{\alpha}/\lambda_{p} \), and knowing that \( \theta_p = 15^{\circ} \), we solve for \( \theta_{\alpha} \) with \( \lambda_{\alpha} = \frac{\lambda_{p}}{\sqrt{2}} \), given the greater mass of alpha particles.
06

Solving for Angle Expression and Numerical Calculation

Given \( \frac{\sin(\theta_{\alpha})}{\sin(15^{\circ})} = \frac{\lambda_{\alpha}}{\lambda_{p}} \) and knowing that \( \sin(\theta_{\alpha}) \approx \frac{\lambda_{p}}{\sqrt{2} \lambda_{p} } \times \sin(15^{\circ}) \), we solve to find that \( \theta_{\alpha} \approx 10.61^{\circ} \).
07

Checking the Calculation

Verify that the calculated angle for the alpha particles results in the expected conditions, confirming that the assumptions about mass and charge ratios lead to consistent application of de Broglie's formula and diffraction.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Proton Beam
A proton beam consists of protons, which are positively charged particles found in the nucleus of an atom. When a beam of protons is accelerated, it gains kinetic energy and can travel at high speeds. In physics, studying a proton beam is important for understanding phenomena such as diffraction. Diffraction occurs when waves encounter an obstacle or pass through a narrow opening.
Protons, due to their mass and charge, exhibit wave-like properties described by quantum mechanics. This is where the concept of de Broglie wavelength comes into play. When a proton beam passes through a small circular hole, it diffracts, forming a pattern of light and dark regions. The dark regions occur at certain angles based on the properties of the proton, the hole, and the wavelength. Understanding this behavior is crucial for applications in fields like materials science and medical imaging.
Alpha Particles
Alpha particles are a type of ionizing radiation composed of two protons and two neutrons, essentially a helium nucleus. They are positively charged and heavier than protons, with a mass of approximately \(6.64 \times 10^{-27} \text{kg}\). Due to their mass and charge, alpha particles carry more energy compared to protons when accelerated by the same electric potential difference. This makes them key players in experiments examining particle diffraction.
In terms of their wave-like behavior, when alpha particles pass through a small opening, they diffract similarly to protons but result in different diffraction patterns because of their larger mass and charge. Observing how the diffraction pattern changes depending on particle mass helps scientists refine our understanding of quantum mechanics and particle interactions.
Potential Difference
Potential difference, often referred to as voltage, is the work needed to move a charge between two points in an electric field. This potential difference accelerates charged particles like protons and alpha particles by giving them kinetic energy. You can express this kinetic energy using the formula:\[ KE = qV, \]where \(q\) is the charge and \(V\) is the potential difference. When equal potential differences are applied to a proton beam and a beam of alpha particles, despite differences in mass and charge, both beams will be accelerated to the same kinetic energy.
This commonality simplifies comparisons in experiments involving particle beams, such as diffraction studies, where scientists analyze how different particles behave under similar energy conditions. Understanding potential difference and its effects on particle motion is crucial in fields like electrical engineering and nuclear physics.
de Broglie Wavelength
The de Broglie wavelength gives a measure of the wave characteristics of particles. It is calculated using the formula:\[ \lambda = \frac{h}{\sqrt{2mKE}}, \]where \(\lambda\) is the de Broglie wavelength, \(h\) is Planck's constant, \(m\) is mass, and \(KE\) is kinetic energy.This concept is essential in quantum mechanics, particularly when analyzing experiments involving diffraction of particles like protons and alpha particles. Since these particles have mass, they exhibit wavelengths according to de Broglie's hypothesis. The wavelength determines the diffraction pattern when particles pass through a small aperture.By examining the resulting diffraction angles, scientists can infer properties about the particles and the structures they pass through. Understanding de Broglie wavelength is pivotal for advances in quantum physics, influencing everything from electron microscopy to semiconductor design.

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Most popular questions from this chapter

A \(10.0-\mathrm{g}\) marble is gently placed on a horizontal tabletop that is 1.75 \(\mathrm{m}\) wide. (a) What is the maximum uncertainty in the horizontal position of the marble? (b) According to the Heisenberg uncertainty principle, what is the minimum uncertainty in the horizontal velocity of the marble? (c) In light of your answer to part (b), what is the longest time the marble could remain on the table? Compare this time to the age of the universe, which is approximately 14 billion years. (Hint: Can you know that the horizontal velocity of the marble is exactly zero?)

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Electrons go through a single slit 150 \(\mathrm{nm}\) wide and strike a screen 24.0 \(\mathrm{cm}\) away. You find that at angles of \(\pm 20.0^{\circ}\) from the center of the diffraction pattern, no electrons hit the screen but electrons hit at all points closer to the center, (a) How fast were these electrons moving when they went through the slit? (b) What will be the next larger angles at which no electrons hit the screen?

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