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A stream of protons, each with a speed of \(0.9900 c\), are directed into a two- slit experiment where the slit separation is \(4.00 \times\) \(10^{-9} \mathrm{~m}\). A two-slit interference pattern is built up on the viewing screen. What is the angle between the center of the pattern and the second minimum (to either side of the center)?

Short Answer

Expert verified
The angle \(\theta\) for the second minimum is found using relativistic adjustments and interference equations.

Step by step solution

01

Understanding the Problem

We are tasked with finding the angle at which the second minimum occurs in a two-slit interference pattern for protons traveling at a relativistic speed of \(0.9900c\). The separation distance between the slits is given as \(4.00 \times 10^{-9} \text{ m}\). This involves applying the formula for constructive or destructive interference.
02

Identify the Relevant Formula for Two-slit Interference

For minima (destructive interference) in a two-slit interference pattern, we use the formula \( d \sin \theta = (m + \frac{1}{2}) \lambda \), where \(d\) is the slit separation, \(\theta\) is the angle of the minimum, \(m\) is the order of the minimum (an integer), and \(\lambda\) is the wavelength of the particles.
03

Calculate De Broglie Wavelength of Protons

Since protons are moving at relativistic speeds, use the relativistic formula for de Broglie wavelength: \( \lambda = \frac{h}{p} \), where \( p = \gamma mv \) is the relativistic momentum, \( h \) is Planck's constant, \( m \) is the proton's rest mass, and \( v = 0.9900c \) is the velocity. Calculate \( \gamma \) using \( \gamma = \frac{1}{\sqrt{1-(v^2/c^2)}} \). Finally, compute \( \lambda \).
04

Apply the Two-slit Interference Formula

With \(m = 1\) for the second minimum, substitute the known values into the formula for minima: \( d \sin \theta = (m + \frac{1}{2}) \lambda \). Solve for \( \theta \) using the calculated de Broglie wavelength, \( d = 4.00 \times 10^{-9} \text{ m} \).
05

Solve for the Angle \( \theta \)

Rearrange the formula to solve for \( \theta \): \( \sin \theta = \frac{(m + \frac{1}{2}) \lambda}{d} \). Calculate \( \sin^{-1} \) (inverse sine) of the result to find the angle \( \theta \).

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Key Concepts

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

Relativistic Physics
Relativistic physics is crucial when dealing with particles moving at speeds close to that of light, represented by \( c \). In our exercise, the protons move at \( 0.9900c \), making relativistic equations essential for accurate calculations.

Unlike classical physics, which can often overlook speed effects at small velocities, relativistic physics incorporates the Lorentz factor (\( \gamma \)), which adjusts for the increase in mass and energy as particles approach the speed of light. This factor is calculated using the formula:

  • \[ \gamma = \frac{1}{\sqrt{1-(v^2/c^2)}} \] where \( v \) is the velocity of the particle.
In essence, as velocity increases towards the speed of light, so does the relativistic mass. Thus, it's imperative to apply relativistic physics to find the true momentum (\( p = \gamma mv \)) and energy of the protons. This adjustment allows us to work with values that reflect reality under extreme conditions.

Understanding these basics provides a foundation to compute accurate results when conducting experiments involving high-speed particles, such as determining interference patterns in a two-slit experiment.
De Broglie Wavelength
The concept of the de Broglie wavelength is fundamental to understanding wave-particle duality. Proposed by Louis de Broglie, it suggests that every moving particle has a wavelength associated with it, reinforcing the idea that particles can exhibit both wave and particle characteristics.

In this scenario with protons, calculating the de Broglie wavelength involves the formula:

  • \[ \lambda = \frac{h}{p} \]
where \( \lambda \) is the de Broglie wavelength, \( h \) is Planck's constant, and \( p \) is momentum. Since the protons are moving at relativistic speeds, their momentum \( p \) is given by \( p = \gamma mv \).

Understanding the de Broglie wavelength helps in calculating the angle of minima in a two-slit interference pattern accurately. The smaller the wavelength, the finer the interference pattern becomes, highlighting the wave-like nature of protons even at atomic scales.

This principle underscores many quantum mechanical phenomena, where particles do not behave as traditional macroscopic objects, but rather as entities with wave properties that are essential for predictions in quantum physics.
Destructive Interference
Destructive interference occurs when waves combine in such a way that they cancel each other out, leading to minima or dark spots in a two-slit interference pattern. When particles like protons interfere destructively, it results in these notable dark areas on the viewing screen.

This phenomenon is described by the equation for minima:

  • \[ d \sin \theta = (m + \frac{1}{2}) \lambda \]
where \( d \) is the slit separation, \( \theta \) is the angle of the minimum, \( m \) is the order of the minimum, and \( \lambda \) is the de Broglie wavelength of the protons.

To find the second minimum, set \( m = 1 \), and solve for \( \theta \) using the previously calculated de Broglie wavelength. By applying the formula above, the interference pattern builds on these dark bands where waves destructively interfere.

This behavior is a key component in wave mechanics and plays a vital role in fields ranging from optics to the quantum study of particles, providing insights into both wave behavior and the inherent nature of particles at the quantum level.

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Most popular questions from this chapter

A light detector (your eye) has an area of \(2.00 \times 10^{-6} \mathrm{~m}^{2}\) and absorbs \(80 \%\) of the incident light, which is at wavelength \(500 \mathrm{~nm}\). The detector faces an isotropic source, \(3.00 \mathrm{~m}\) from the source. If the detector absorbs photons at the rate of exactly \(4.000 \mathrm{~s}^{-1}\), at what power does the emitter emit light?

An orbiting satellite can become charged by the photoelectric effect when sunlight ejects electrons from its outer surface. Satellites must be designed to minimize such charging because it can ruin the sensitive microelectronics. Suppose a satellite is coated with platinum, a metal with a very large work function \((\Phi=5.32 \mathrm{eV})\). Find the longest wavelength of incident sunlight that can eject an electron from the platinum.

what is the maximum kinetic energy of electrons knocked out of a thin copper foil by Compton scattering of an incident beam of \(17.5 \mathrm{keV}\) x rays? Assume the work function is negligible.

You will find in Chapter 39 that electrons cannot move in definite orbits within atoms, like the planets in our solar system. To see why, let us try to "observe" such an orbiting electron by using a light microscope to measure the electron's presumed orbital position with a precision of, say, \(10 \mathrm{pm}\) (a typical atom has a radius of about \(100 \mathrm{pm}\) ). The wavelength of the light used in the microscope must then be about \(10 \mathrm{pm}\). (a) What would be the photon energy of this light? (b) How much energy would such a photon impart to an electron in a head-on collision? (c) What do these results tell you about the possibility of "viewing" an atomic electron at two or more points along its presumed orbital path? (Hint: The outer electrons of atoms are bound to the atom by energies of only a few electron-volts.)

A 3.0 MeV proton is incident on a potential energy barrier of thickness \(10 \mathrm{fm}\) and height \(10 \mathrm{MeV}\). What are (a) the transmission coefficient \(T,(\mathrm{~b})\) the kinetic energy \(K_{t}\) the proton will have on the other side of the barrier if it tunnels through the barrier, and (c) the kinetic energy \(K_{r}\) it will have if it reflects from the barrier? A \(3.0 \mathrm{MeV}\) deuteron (the same charge but twice the mass as a proton) is incident on the same barrier. What are (d) \(T\), (e) \(K_{t}\), and (f) \(K_{r}\) ?

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