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Doorway diffraction. If your wavelength were \(1.0 \mathrm{m},\) you would undergo considerable diffraction in moving through a doorway. (a) What must your speed be for you to have this wavelength? (Assume that your mass is 60.0 \(\mathrm{kg} .\) ) (b) At the speed calculated in part (a), how many years would it take you to move 0.80 \(\mathrm{m}\) (one step)? Will you notice diffraction effects as you walk through doorways?

Short Answer

Expert verified
Speed: \(1.104 \times 10^{-35} \, \mathrm{m/s}\); Time: \(2.30 \times 10^{27} \) years. Diffraction is not noticeable.

Step by step solution

01

Understanding the Problem

We need to calculate the speed at which an object (in this case, a person with mass 60.0 kg) has a wavelength of 1.0 m using the de Broglie wavelength formula. Then, determine the time it takes to move a short distance at this speed.
02

Apply de Broglie Wavelength Formula

The de Broglie wavelength formula is given by: \[ \lambda = \frac{h}{mv} \]where \( \lambda = 1.0 \, \mathrm{m} \), \( m = 60.0 \, \mathrm{kg} \), and \( h = 6.626 \times 10^{-34} \, \mathrm{m^2 \, kg \, s^{-1}} \). Rearrange to solve for speed \( v \):\[ v = \frac{h}{m\lambda} = \frac{6.626 \times 10^{-34}}{60.0 \times 1.0} \approx 1.104 \times 10^{-35} \, \mathrm{m/s} \].
03

Calculate Time to Move 0.80 m

Use the formula for time, \( t = \frac{distance}{speed} \), where the distance is 0.80 m. Substitute the speed calculated from step 2:\[ t = \frac{0.80}{1.104 \times 10^{-35}} \approx 7.25 \times 10^{34} \, \mathrm{s} \].
04

Convert Time to Years

Convert the time calculated from seconds to years using:\( 1 \, \text{year} = 3.154 \times 10^{7} \, \text{seconds} \).\[ \text{Time in years} = \frac{7.25 \times 10^{34}}{3.154 \times 10^{7}} \approx 2.30 \times 10^{27} \, \text{years} \].
05

Conclusion on Diffraction Effects

At the calculated speed, the time to move just one step is far longer than the age of the universe. Thus, no diffraction effects are observable as you walk through doorways because such a speed is impractical and irrelevant for everyday observations.

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Key Concepts

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

Quantum Mechanics
Quantum mechanics is the branch of physics that deals with the behavior of very tiny particles, like electrons and photons, on the atomic and subatomic scales. Unlike classical physics, which describes everyday objects and their behaviors in predictable ways, quantum mechanics introduces us to a somewhat different world. In this world, particles can exist in multiple states at once and can also behave like waves.

One of the central ideas in quantum mechanics is that particles have both wave-like and particle-like properties, which leads to interesting phenomena such as diffraction. Diffraction is the bending of waves around obstacles or openings and is a phenomenon that can be observed with light, sound, and other waves. In the exercise above, we're introduced to the idea of 'doorway diffraction' with a wavelength of 1 meter, which is an unusual scenario for human-sized objects.

In quantum mechanics, we also encounter uncertainty. For instance, the more precisely we know the position of a particle, the less precisely we can know its momentum, and vice versa. This is known as Heisenberg's Uncertainty Principle, and it profoundly affects how we understand particle interactions on a very small scale.
Wave-Particle Duality
Wave-particle duality is a core concept in quantum mechanics that describes how particles like electrons and light can exhibit both wave-like and particle-like properties. This duality challenges classical intuition, which tends to separate particles and waves into distinct categories.

Albert Einstein and others found that light, which behaves like a wave (as seen in patterns like interference and diffraction), also displays particle-like characteristics. Light can be thought of as being made up of particles called photons. Similarly, de Broglie's hypothesis proposed that matter can also behave like waves, with associated wavelengths known as de Broglie wavelengths.

The de Broglie wavelength ( \( \lambda \)) describes the wave nature of a particle and is given by the formula:
- \( \lambda = \frac{h}{mv} \)
where \( h \) is Planck's constant, \( m \) is the particle's mass, and \( v \) is its velocity.

In our exercise, we explored the de Broglie wavelength for a human moving through a doorway. Even though it's a theoretical exploration, the calculations show how wave-particle duality extends beyond subatomic particles to all matter, even if practically, we don't observe it in large objects like humans.
Physics Problem Solving
Physics problem solving involves understanding the problem, applying relevant formulas, manipulating mathematical expressions, and interpreting the results. In the exercise, we start by identifying the goal: calculating the speed needed for a human with a given mass to have a de Broglie wavelength of 1 meter.

The approach involves these steps:
  • Understanding what is being asked: We must first comprehend the concept of de Broglie wavelength and how it's calculated.
  • Choosing the right formula: In this case, using \( \lambda = \frac{h}{mv} \).
  • Rearranging the formula: To solve for \( v \), rearranging gives \( v = \frac{h}{m\lambda} \).
  • Substituting given values: We plug in the given mass and wavelength to find the speed.
  • Calculating related concepts: Finally, determining how long it would take to move a particular distance at that speed.
Problem solving in physics is not just about crunching numbers, but about connecting mathematical results with real-world concepts. The ultimate conclusion of the exercise was that the calculated speed and time indicate why quantum effects like diffraction are not noticeable in everyday human experiences.

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Most popular questions from this chapter

(a) What is the least amount of energy, in electron volts, that must be given to a hydrogen atom which is initially in its ground level so that it can emit the \(\mathrm{H}_{\alpha}\) line in the Balmer series? (b) How many different possibilities of spectral-line emissions are there for this atom when the electron starts in the \(n=3\) level and eventually ends up in the ground level? Calculate the wavelength of the emitted photon in each case.

\(\bullet\) A hydrogen atom initially in the ground state absorbs a photon, which excites it to the \(n=4\) state. Determine the wavelength and frequency of the photon.

\(\bullet\) (a) The uncertainty in the \(x\) component of the position of a proton is \(2.0 \times 10^{-12} \mathrm{m}\) . What is the minimum uncertainty in the \(x\) component of the velocity of the proton? (b) The uncertainty in the \(x\) component of the velocity of an electron is 0.250 \(\mathrm{m} / \mathrm{s} .\) What is the minimum uncertainty in the \(x\) coordinate of the electron?

If a photon of wavelength 0.04250 nm strikes a free electron and is scattered at an angle of \(35.0^{\circ}\) from its original direction, find (a) the change in the wavelength of this photon, (b) the wavelength of the scattered light, (c) the change in energy of the photon (is it a loss or a gain?), and (d) the energy gained by the electron.

\(\bullet\) The photoelectric threshold wavelength of a tungsten surface is 272 \(\mathrm{nm}\) . (a) What are the threshold frequency and work function (in eV) of this tungsten? (b) Calculate the maximum kinetic energy (in eV) of the electrons ejected from this tungsten surface by ultraviolet radiation of frequency \(1.45 \times 10^{15} \mathrm{Hz}\)

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