Problem 69 - Chemistry: An Atoms-Focused Approach

Question

Calculate the wavelengths of the following objects: a. A muon (a subatomic particle with a mass of \(1.884 \times\) \(\left.10^{-28} \mathrm{kg}\right)\) traveling at \(325 \mathrm{m} / \mathrm{s}\) b. Electrons \(\left(m_{c}=9.10938 \times 10^{-31} \mathrm{kg}\right)\) moving at \(4.05 \times\) \(10^{6} \mathrm{m} / \mathrm{s}\) in an electron microscope c. An \(82 \mathrm{kg}\) sprinter running at \(9.9 \mathrm{m} / \mathrm{s}\) d. Earth (mass \(\left.=6.0 \times 10^{24} \mathrm{kg}\right)\) moving through space at \(3.0 \times 10^{4} \mathrm{m} / \mathrm{s}\)

Step-by-Step Solution

Answer

Question: Calculate the wavelengths of the following objects: a. A muon with mass 1.884 x 10^-28 kg and velocity 325 m/s b. An electron with mass 9.10938 x 10^-31 kg and velocity 4.05 x 10^6 m/s c. A sprinter with mass 82 kg and velocity 9.9 m/s d. The Earth with mass 6.0 x 10^24 kg and velocity 3.0 x 10^4 m/s Answer: a. Muon wavelength: 1.082 x 10^-8 m b. Electron wavelength: 1.797 x 10^-10 m c. Sprinter wavelength: 8.157 x 10^-37 m d. Earth wavelength: 3.681 x 10^-63 m

Step 1: Identify the muon's mass and velocity

The mass of the muon is given as \(1.884 \times 10^{-28} \mathrm{kg}\) and its velocity is \(325 \mathrm{m/s}\).

Step 2: Calculate the muon's momentum

Using the formula \(p = mv\), where \(m = 1.884 \times 10^{-28} \mathrm{kg}\) and \(v = 325 \mathrm{m/s}\), we get: \(p = (1.884 \times 10^{-28} \mathrm{kg})(325 \mathrm{m/s}) = 6.123 \times 10^{-26} \mathrm{kg\:m/s}\).

Step 3: Calculate the muon's wavelength

Using the de Broglie wavelength formula, \(\lambda = \frac{h}{p}\): \(\lambda_\text{muon} = \frac{6.626 \times 10^{-34} \mathrm{Js}}{6.123 \times 10^{-26} \mathrm{kg\:m/s}} = 1.082 \times 10^{-8} \mathrm{m}\) #b. Electron Wavelength#

Step 1: Identify the electron's mass and velocity

The mass of the electron is given as \(9.10938 \times 10^{-31} \mathrm{kg}\) and its velocity is \(4.05 \times 10^6 \mathrm{m/s}\).

Step 2: Calculate the electron's momentum

Using the formula \(p = mv\), where \(m = 9.10938 \times 10^{-31} \mathrm{kg}\) and \(v = 4.05 \times 10^6 \mathrm{m/s}\), we get: \(p = (9.10938 \times 10^{-31} \mathrm{kg})(4.05 \times 10^6 \mathrm{m/s}) = 3.689 \times 10^{-24} \mathrm{kg\: m/s}\).

Step 3: Calculate the electron's wavelength

Using the de Broglie wavelength formula, \(\lambda = \frac{h}{p}\): \(\lambda_\text{electron} = \frac{6.626 \times 10^{-34} \mathrm{Js}}{3.689 \times 10^{-24} \mathrm{kg\: m/s}} = 1.797 \times 10^{-10} \mathrm{m}\) #c. Sprinter Wavelength#

Step 1: Identify the sprinter's mass and velocity

The mass of the sprinter is given as \(82 \mathrm{kg}\) and his velocity is \(9.9 \mathrm{m/s}\).

Step 2: Calculate the sprinter's momentum

Using the formula \(p = mv\), where \(m = 82 \mathrm{kg}\) and \(v = 9.9 \mathrm{m/s}\), we get: \(p = (82 \mathrm{kg})(9.9 \mathrm{m/s}) = 812.1 \mathrm{kg\: m/s}\).

Step 3: Calculate the sprinter's wavelength

Using the de Broglie wavelength formula, \(\lambda = \frac{h}{p}\): \(\lambda_\text{sprinter} = \frac{6.626 \times 10^{-34} \mathrm{Js}}{812.1 \mathrm{kg\: m/s}} = 8.157 \times 10^{-37} \mathrm{m}\) #d. Earth Wavelength#

Step 1: Identify the Earth's mass and velocity

The mass of the Earth is given as \(6.0 \times 10^{24} \mathrm{kg}\) and its velocity is \(3.0 \times 10^{4} \mathrm{m/s}\).

Step 2: Calculate Earth's momentum

Using the formula \(p = mv\), where \(m = 6.0 \times 10^{24} \mathrm{kg}\) and \(v = 3.0 \times 10^{4} \mathrm{m/s}\), we get: \(p = (6.0 \times 10^{24} \mathrm{kg})(3.0 \times 10^{4} \mathrm{m/s}) = 1.8 \times 10^{29} \mathrm{kg\: m/s}\).

Step 3: Calculate the Earth's wavelength

Using the de Broglie wavelength formula, \(\lambda = \frac{h}{p}\): \(\lambda_\text{Earth} = \frac{6.626 \times 10^{-34} \mathrm{Js}}{1.8 \times 10^{29} \mathrm{kg\: m/s}} = 3.681 \times 10^{-63} \mathrm{m}\)

Key Concepts

Wave-Particle Duality

Wave-particle duality is a fundamental concept in quantum mechanics. It describes how every particle or quantum entity may exhibit both wave-like and particle-like properties. This duality is most famously demonstrated in particles like electrons and photons, which sometimes behave like waves (interference and diffraction) and sometimes like particles (collision and scattering).

The de Broglie hypothesis is pivotal to understanding this duality. Proposed by Louis de Broglie in 1924, it suggests that particles can have a wavelength \[ \lambda = \frac{h}{p} \] where \(\lambda\) is the wavelength, \(h\) is Planck's constant \(6.626 \times 10^{-34} \, \mathrm{Js}\), and \(p\) is momentum. This formula links wave nature and particle nature together by showing that movement and mass contribute to wavelength properties.

This duality plays a critical role not only in subatomic particles but in all scales of nature, influencing fields such as optics and materials science.

Momentum Calculation

Momentum is an essential concept in physics, quantifying the motion of an object. It's a vector quantity, meaning it has both magnitude and direction. The formula for linear momentum is \( p = mv \), where \(m\) is mass and \(v\) is velocity.

Momentum is conserved in closed systems, and understanding it is crucial for calculating objects' behavior during interactions or collisions. In the problem discussed, calculating momentum is the first step to finding the de Broglie wavelength of various entities, such as a muon, electron, sprinter, and Earth.

Key points about momentum:

• It shows how massive and fast an object is moving.
• Higher momentum means a bigger impact when the object interacts with another.
• It helps in understanding the relationship between energy and motion.

Momentum calculations are fundamental, bridging classical and modern physics concepts through the de Broglie wavelength.

Subatomic Particle Physics

Subatomic particle physics examines the fundamental particles of the universe and their interactions. These include particles like electrons, protons, neutrons, and more exotic ones like muons. Every subatomic particle has unique properties like mass, charge, and spin, contributing to the complexity of atomic and nuclear physics.

Within this realm, muons and electrons are interesting due to their behavior and properties:

• Muons are similar to electrons but are heavier, making them more unstable.
• Electrons, as one of the basic constituents of atoms, are fundamental to understanding chemical bonds and reactions.

These particles demonstrate wave-particle duality prominently.

Subatomic physics relies heavily on experiments and mathematical models to understand phenomena. From electron microscopes using electron wavelengths to particle accelerators revealing new particles, subatomic physics continues to explore the universe's building blocks, explaining the fundamental forces and interactions at their most basic level.