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The mass of a particle is \(10^{-10} \mathrm{~g}\) and its diameter is \(10^{-4} \mathrm{~cm} .\) If its speed is \(10^{-6} \mathrm{~cm} / \mathrm{s}\) with \(0.0001 \%\) uncertainty in measurement, the minimum uncertainty in its position is (a) \(5.28 \times 10^{-8} \mathrm{~m}\) (b) \(5.28 \times 10^{-7} \mathrm{~m}\) (c) \(5.28 \times 10^{-6} \mathrm{~m}\) (d) \(5.28 \times 10^{-9} \mathrm{~m}\)

Short Answer

Expert verified
The minimum uncertainty in its position is \(5.28 \times 10^{-8} \mathrm{~m}\).

Step by step solution

01

Understand the Heisenberg's Uncertainty Principle

Heisenberg's Uncertainty Principle can be written as \( \Delta x \cdot \Delta p \geq \frac{\hbar}{2} \), where \( \Delta x \) is the uncertainty in position, \( \Delta p \) is the uncertainty in momentum, and \( \hbar \) is the reduced Planck's constant equal to \(1.0545718 \times 10^{-34} \textrm{Js}\). This principle sets a limit to the precision with which position \(x\) and momentum \(p\) can be simultaneously known. The task is to calculate the minimum uncertainty in position, knowing the uncertainty in speed.
02

Calculate Uncertainty in Speed

Given the speed of the particle is \(1 \times 10^{-6} \textrm{cm/s}\) and the uncertainty in speed measurement is \(0.0001\%\). To find the uncertainty in speed, multiply the speed by the percentage uncertainty: \(\Delta v = v \times \text{percentage uncertainty} = 10^{-6} \textrm{cm/s} \times 0.0001/100 = 10^{-12} \textrm{cm/s}\).
03

Convert the Uncertainty in Speed to SI Units

To work within the SI system, convert \(10^{-12} \textrm{cm/s}\) to meters per second: \(\Delta v = 10^{-12} \textrm{cm/s} \times 10^{-2} \textrm{m/cm} = 10^{-14} \textrm{m/s}\).
04

Calculate the Uncertainty in Momentum

The uncertainty in momentum (\( \Delta p\)) can be found using the particle's mass (\( m \) in kilograms) and the uncertainty in velocity (\( \Delta v \)), as \( \Delta p = m \cdot \Delta v\). Convert the mass to kilograms: \(10^{-10} \textrm{g} = 10^{-13} \textrm{kg}\). Now calculate: \( \Delta p = 10^{-13} \textrm{kg} \times 10^{-14} \textrm{m/s} = 10^{-27} \textrm{kg} \cdot \textrm{m/s}\).
05

Apply the Uncertainty Principle to Find the Minimum Uncertainty in Position

Use the uncertainty principle to find \( \Delta x\): \( \Delta x \cdot \Delta p \geq \frac{\hbar}{2} \). Solving for \( \Delta x \) gives \( \Delta x \geq \frac{\hbar}{2 \cdot \Delta p} \). Plugging in the values: \( \Delta x \geq \frac{1.0545718 \times 10^{-34}}{2 \times 10^{-27}} \approx 5.28 \times 10^{-8} \textrm{m}\).

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Key Concepts

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

Uncertainty in Position
The uncertainty in position, denoted by \( \Delta x \), is a key concept in quantum mechanics reflecting the idea that the exact position of a particle cannot be determined with absolute precision. This principle challenges our classical intuition, which suggests that objects should have definite locations. In the context of the given exercise, the minimum uncertainty in position of a particle is linked to the uncertainty in its momentum through Heisenberg's Uncertainty Principle. The principle asserts that as the precision in measuring the momentum of a particle increases, the ability to pinpoint its position diminishes, and vice versa.

To make this concept easier to digest, imagine trying to measure a rapidly moving object with blurry vision; the faster it moves, the harder it is to say exactly where it is. Similarly, in quantum mechanics, particles such as electrons behave in ways where their exact locations are inherently vague. This vagueness is not due to the limits of our instruments, but due to the nature of reality at a quantum level. The exercise demonstrates this by calculating the minimum uncertainty in position given an uncertainty in the particle's speed, using the formula derived from the uncertainty principle.
Uncertainty in Momentum
Uncertainty in momentum, expressed as \( \Delta p \), is the measure of how much the momentum of a particle is unknown. It complements the uncertainty in position when considering the behavior of quantum particles. The momentum of a particle is the product of its mass and velocity, and in quantum mechanics, it's impossible to know this value with perfect accuracy.

The step by step solution outlined in the exercise illustrates how to calculate the uncertainty in momentum, which is pivotal in determining the corresponding uncertainty in position using the Heisenberg's Uncertainty Principle. To calculate \( \Delta p \), the exercise begins with the known mass of a particle and an uncertainty in its speed, then combines these to find the product—representing the uncertainty in momentum. It is essential in physical chemistry, especially for competitive examinations, to understand this process and be able to apply the principle to a variety of scenarios involving subatomic particles.
Quantum Mechanics
Quantum mechanics is a fundamental theory in physics that provides a description of the physical properties of nature at the scale of atoms and subatomic particles. It's a complex field that uses mathematical concepts and principles to explain phenomena that cannot be observed directly, including the behavior of particles and their interactions.

The Heisenberg's Uncertainty Principle is a cornerstone of quantum mechanics. It teaches us that certain pairs of physical properties, like position and momentum, cannot be measured precisely at the same time. Quantum mechanics discards the determinism of classical mechanics; instead, it introduces probability and uncertainty into our understanding of the microworld. The calculations present in the exercise are a straightforward application of quantum principles, helping students familiarize themselves with the mathematical approach used in this field. These concepts are integral to advanced studies in physical chemistry and are often the subjects of questions in competitive examinations.
Physical Chemistry for Competitive Examinations
In the realm of physical chemistry for competitive examinations, understanding and applying concepts like the Heisenberg's Uncertainty Principle is essential. These concepts are frequently tested in exams that assess a student's readiness for further education in science and technology fields. As an example, questions might require calculating the uncertainty in position or momentum of a particle based on provided data.

Such examinations test a student's grasp of theory, as well as their ability to solve problems and think critically about physical phenomena at the quantum level. The exercise we've explored serves as an excellent example of the type of problem-solving skills required. Students must not only memorize the formulas but also understand the underlying principles and be able to apply them to different situations. By mastering these topics, students can significantly enhance their performance in competitive examinations within the sphere of physical chemistry.

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Most popular questions from this chapter

Correct set of quantum numbers defining the highest energy electron in scandium (I) ion is (a) \(n=3, l=1, m=0, s=-1 / 2\) (b) \(n=3, l=0, m=0, s=-1 / 2\) (c) \(n=4, l=0, m=0, s=+1 / 2\) (d) \(n=3, l=2, m=2, s=+1 / 2\)

Photoelectrons are liberated by ultraviolet light of wavelength \(3000 \AA\) from a metallic surface for which the photoelectric threshold is \(4000 \AA\). The de-Broglie wavelength of electrons emitted with maximum kinetic energy is (a) \(1000 \AA\) (b) \(42.43 \AA\) (c) \(12.05 \AA\) (d) \(3.54 \AA\)

The ratio of de-Broglie wavelength of electron and proton moving with the same speed is about (a) \(1836: 1\) (b) \(1: 1836\) (c) \(1: 1\) (d) \(1: 2\)

For an electron in a hydrogen atom, the wave function is given by \(\psi_{1 \mathrm{~s}}\) \(=(\pi / \sqrt{2}) e^{-r / a_{0}}\), where \(a_{0}\) is the radius of first Bohr's orbit and \(r\) is the distance from the nucleus with which probability of finding electron varies. What will be the ratio of probabilities of finding electrons at the nucleus to first Bohr's orbit \(a_{\mathrm{o}} ?\) (a) 0 (b) \(e\) (c) \(e^{2}\) (d) \(\frac{1}{e^{2}}\)

An electron at rest is accelerated through a potential difference of \(200 \mathrm{~V}\). If the specific charge of electron is \(1.76 \times 10^{11} \mathrm{C} / \mathrm{kg}\), the speed acquired by the electron is about (a) \(8.4 \times 10^{6} \mathrm{~cm} / \mathrm{s}\) (b) \(8.4 \times 10^{6} \mathrm{~m} / \mathrm{s}\) (c) \(4.2 \times 10^{6} \mathrm{~m} / \mathrm{s}\) (d) \(4.2 \times 10^{6} \mathrm{~cm} / \mathrm{s}\)

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