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A scientist has devised a new method of isolating individual particles. He claims that this method enables him to detect simultaneously the position of a particle along an axis with a standard deviation of 0.12 \(\mathrm{nm}\) and its momentum component along this axis with a standard deviation of \(3.0 \times 10^{-25} \mathrm{kg} \cdot \mathrm{m} / \mathrm{s} .\) Use the Heisenberg uncertainty principle to evaluate the validity of this claim.

Short Answer

Expert verified
The claim violates the Heisenberg uncertainty principle.

Step by step solution

01

Understanding Heisenberg's Uncertainty Principle

The Heisenberg uncertainty principle states that the product of the uncertainties in position (abla x) and momentum (abla p) of a particle is always greater than or equal to \( \frac{\hbar}{2} \), where \( \hbar \) is the reduced Planck's constant (\( \hbar = 1.05 \times 10^{-34} \mathrm{J} \cdot \mathrm{s} \)). Mathematically, this can be represented as: \( abla x \cdot abla p \geq \frac{\hbar}{2} \).
02

Calculate the Product of Uncertainties

Using the given standard deviations, \( abla x = 0.12 \mathrm{nm} = 0.12 \times 10^{-9} \mathrm{m} \) and \( abla p = 3.0 \times 10^{-25} \mathrm{kg} \cdot \mathrm{m} / \mathrm{s} \), calculate the product: \( abla x \cdot abla p = (0.12 \times 10^{-9}) \cdot (3.0 \times 10^{-25}) \mathrm{kg} \cdot \mathrm{m} \cdot \mathrm{s}^{-1} \).
03

Compare Product to Reduced Planck's Constant

Calculate \( \frac{\hbar}{2} = \frac{1.05 \times 10^{-34}}{2} \) J\cdot s. Then compare this value to the product of the uncertainties calculated in Step 2. If \( abla x \cdot abla p < \frac{\hbar}{2} \), the claim violates the uncertainty principle.
04

Evaluation

Substituting the values, we have \( abla x \cdot abla p = (0.12 \times 10^{-9}) \cdot (3.0 \times 10^{-25}) = 3.6 \times 10^{-35} \mathrm{kg} \cdot \mathrm{m}^2 \cdot \mathrm{s}^{-1} \). Calculate \( \frac{\hbar}{2} = 0.525 \times 10^{-34} \approx 5.25 \times 10^{-35} \) J\cdot s. \( 3.6 \times 10^{-35} \mathrm{kg} \cdot \mathrm{m}^2 \cdot \mathrm{s}^{-1} < 5.25 \times 10^{-35} \mathrm{kg} \cdot \mathrm{m}^2 \cdot \mathrm{s}^{-1} \), which violates the uncertainty principle.

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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 explores the behavior and interactions of particles at the smallest scales—atoms and subatomic particles. It challenges our classical understanding by introducing concepts that can seem counterintuitive at first. Unlike classical physics, where objects have definite properties like position and speed, quantum mechanics proposes that on a microscopic scale, particles exist in a state of probability.

At this level, we speak of wave functions that describe the probabilities of finding a particle in a particular state. However, when measurements are taken, these states "collapse" to specific outcomes. This fascinating and bewildering world is governed by principles including the superposition of states, wave-particle duality, and of course, the celebrated Heisenberg Uncertainty Principle, which plays a crucial role in defining the limits of our measurement capabilities.
Standard Deviation
In the realm of quantum mechanics, assessing how precisely we know a particle’s characteristics, like its position or momentum, requires statistical measures. Standard deviation is one such measure, indicating the variability or spread of a set of data points. Essentially, it tells us how much the values in a data set differ from the average or mean value.

In our scenario, standard deviation helps quantify the 'uncertainty' or 'spread' in the potential measurements of a particle's position and momentum. For example, a standard deviation of 0.12 nm for position suggests that the particle could be found anywhere within a relatively small range around this value. Similarly, a momentum with a standard deviation of \(3.0 \times 10^{-25}\) kgâ‹…m/s indicates the uncertainty in its momentum measurement. Thus, standard deviation is pivotal in quantifying uncertainty in quantum experiments.
Reduced Planck's Constant
The reduced Planck's constant, symbolized as \( \hbar \) (h-bar), is a fundamental constant in quantum mechanics. It is derived from the Planck's constant \( h \), divided by \( 2\pi \), and is approximately \( 1.05 \times 10^{-34} \) Jâ‹…s. This constant is crucial in the context of the Heisenberg Uncertainty Principle.

In the Heisenberg Uncertainty Principle, the reduced Planck's constant represents the smallest scale at which the uncertainties in measurements can interact. It underscores limit in precision that can be achieved when measuring pairs of complementary properties, like position and momentum. Consequently, \( \hbar \) plays a central role not just in theoretical formulations but also in our practical understanding of the quantum world.
Position and Momentum Uncertainty
The concept of position and momentum uncertainty is intricately tied to Heisenberg's Uncertainty Principle. This principle posits that the more precisely we determine a particle's position, the less precisely we can know its momentum, and vice versa.

Mathematically, this relationship is expressed as \( \Delta x \cdot \Delta p \geq \frac{\hbar}{2} \), where \( \Delta x \) and \( \Delta p \) denote the standard deviations (uncertainties) in position and momentum, respectively. In simpler terms, it's nature's way of imposing a limit on our knowledge of both position and momentum at the same time.

For instance, when the calculated product of uncertainties in position and momentum is less than \( \frac{\hbar}{2} \), it indicates a violation of the Uncertainty Principle. This principle is fundamental in understanding why precise measurements at such small scales remain elusive and continues to challenge our perception of nature's intrinsic randomness and indeterminacy.

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Most popular questions from this chapter

The negative muon has a charge equal to that of an electron but a mass that is 207 times as great. Consider a hydrogenlike atom consisting of a proton and a muon. (a) What is the reduced mass of the atom? (b) What is the ground-level energy (in electron volts)? (c) What is the wavelength of the radiation emitted in the transition from the \(n=2\) level to the \(n=1\) level?

(a) If a photon and an electron each have the same energy of 20.0 eV, find the wavelength of each. (b) If a photon and anelectron each have the same wavelength of 250 \(\mathrm{nm}\) , find the energy of each. (c) You want to study an organic molecule that is about 250 \(\mathrm{nm}\) long using either a photon or an electron microscope .Approximately what wavelength should you use, and which probe, the electron or the photon, is likely to damage the molecule the least?

A pesky 1.5-mg mosquito is annoying you as you attempt to study physics in your room, which is 5.0 m wide and 2.5 \(\mathrm{m}\) high. You decide to swat the bothersome insect as it flies toward you, but you need to estimate its speed to make a successful hit. (a) What is the maximum uncertainty in the horizontal position of the mosquito? (b) What limit does the Heisenberg uncertainty principle place on your ability to know the horizontal velocity of this mosquito? Is this limitation a serious impediment to your attempt to swat it?

A beam of \(40-\mathrm{eV}\) electrons traveling in the \(+x-\) direction passes through a slit that is parallel to the \(y\) -axis and 5.0\(\mu \mathrm{m}\) wide. The diffraction pattern is recorded on a screen 2.5 \(\mathrm{m}\) from the slit. (a) What is the de Broglie wavelength of the electrons? (b) How much time does it take the electrons to travel from the slit to the screen? (c) Use the width of the central diffraction pattern to calculate the uncertainty in the \(y\) -component of momentum of an electron just after it has passed through the slit. (d) Use the result of part (c) and the Heisenberg uncertainty principle (Eq. 39.29 for \(y\) ) to estimate the minimum uncertainty in the \(y\) -coordinate of an electron just after it has passed through the slit. Compare your result to the width of the slit.

Calculate the energy in electron volts of (a) an electron that has de Broglie wavelength 400 \(\mathrm{nm}\) and (b) a photon that has wavelength 400 \(\mathrm{nm} .\)

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