Question

Consider a molecule of hydrogen at \(478.15 \mathrm{~K}\) and moving with a velocity of \(2 \times 10^{5} \mathrm{~cm} / \mathrm{sec} .\) What is the de Broglie wavelength of this hydrogen atom?

Short answer

The de Broglie wavelength of the hydrogen molecule is approximately \(5.0 \times 10^{-11}\) meters.

Step-by-step solution

Finding the mass of the hydrogen molecule

Since it is a hydrogen molecule, which contains two hydrogen atoms, the mass of the hydrogen molecule is double the mass of a hydrogen atom. The mass of a hydrogen atom is approximately 1 u (unified atomic mass unit), so the mass of a hydrogen molecule is 2 u. We need the mass in kilograms to calculate the momentum. The conversion factor between atomic mass units and kilograms is: 1 u = 1.66054 × 10^{-27} kg Thus, the mass of a hydrogen molecule in kg is: \(2 \times 1.66054 \times 10^{-27} \mathrm{kg}\).

Calculating the momentum of the hydrogen molecule

Now we will calculate the momentum of the hydrogen molecule using the formula: Momentum (p) = Mass (m) × Velocity (v) Using the values from Step 1, the mass of the hydrogen molecule (m) is equal to \(2 \times 1.66054 \times 10^{-27} \mathrm{kg}\), and the given velocity (v) is \(2 \times 10^{5} \mathrm{cm} / \mathrm{sec}\). First, convert the velocity to meters per second: \(2 \times 10^5 \mathrm{cm} / \mathrm{sec} = 2 \times 10^3 \mathrm{m} / \mathrm{sec}\). Now, multiply the mass and velocity to find the momentum: p = \( (2 \times 1.66054 \times 10^{-27} \mathrm{kg}) \times (2 \times 10^3 \mathrm{m} / \mathrm{sec})\)

Finding the de Broglie wavelength

We'll use the de Broglie wavelength formula to calculate the wavelength: \(λ = \frac{h}{p}\) Where λ is the de Broglie wavelength, h is the Planck's constant (\(6.626 \times 10^{-34} \mathrm{Js}\)), and p is the momentum of the particle. Now substitute the momentum calculated in Step 2 into the formula: \(λ = \frac{6.626 \times 10^{-34} \mathrm{Js}}{(2 \times 1.66054 \times 10^{-27} \mathrm{kg}) \times (2 \times 10^3 \mathrm{m} / \mathrm{sec})}\)

Calculating the de Broglie wavelength

Finally, calculate the value for the de Broglie wavelength: \(λ \approx 5.0 \times 10^{-11} \mathrm{m}\) So, the de Broglie wavelength of the hydrogen molecule is approximately \(5.0 \times 10^{-11}\) meters.

Key concepts

Quantum Mechanics

Quantum mechanics is a branch of physics that explains the behavior of particles at the smallest scales, where classical physics no longer provides accurate predictions. One of its essential aspects is that particles, like electrons and even larger objects like molecules, exhibit both wave-like and particle-like properties.
This dual nature leads to interesting phenomena like quantum superposition and entanglement, which challenge our classical understanding of physics.

• Particles, even at rest, have associated wave properties. This is at the core of de Broglie's hypothesis.
• The smaller the particle, the more significant these quantum mechanical effects become.

The de Broglie wavelength is a specific concept in quantum mechanics indicating that every particle of matter can be associated with a wave. This is the crux of what we explore with the hydrogen molecule in the initial exercise.

Hydrogen Molecule

A hydrogen molecule, denoted as H extsubscript{2}, is made up of two hydrogen atoms bonded together. It is the simplest molecule in the universe and is the basic building block of many chemical compounds.
Understanding the properties of hydrogen molecules is crucial in fields like chemistry and astrophysics.

• Each hydrogen atom has a single proton in its nucleus. When two hydrogen atoms combine, they form a molecule that has a total of two protons and is electrically neutral.
• H extsubscript{2} is a diatomic molecule, meaning it's composed of two atoms.

The mass of a hydrogen molecule plays a vital role in quantum mechanical calculations, as seen in the exercise where we calculated the de Broglie wavelength, using the total mass of two hydrogen atoms.

Unified Atomic Mass Unit

The unified atomic mass unit, often abbreviated as u or amu, is a standard unit of mass used to express atomic and molecular weights. It is defined as one twelfth of the mass of an unbound neutral atom of carbon-12 in its nuclear and electronic ground state and is approximately equal to 1.66054 × 10^{-27} kilograms.
This unit allows scientists to compare the mass of atoms and molecules on a relative scale.

• An atomic mass unit provides a convenient way to express the mass of atoms in chemistry and physics.
• It makes it easy to deal with the exceptionally small masses of atoms without resorting to very large negative exponents.

In our exercise, the unified atomic mass unit was crucial in converting the mass of the hydrogen molecule from atomic mass units to kilograms for calculation of momentum.

Momentum Calculation

Momentum is a fundamental concept in physics, describing the quantity of motion an object possesses, given by the product of its mass and velocity. In quantum mechanics, calculating the momentum of particles like molecules helps us derive further properties, such as the de Broglie wavelength.

• The formula for momentum is simple: \[ p = m \times v \]
• Where \( p \) is momentum, \( m \) is mass, and \( v \) is velocity.

This simple formula takes on new significance in quantum mechanics, because momentum is integral to determining characteristics like wave-particle duality of matter.
In the provided exercise, knowing the precise momentum enabled us to calculate the de Broglie wavelength of the hydrogen molecule, using Planck's constant and demonstrating the wave nature of matter at small scales.