Problem 55 - Chemistry An Atoms First Approach

Question

A particle has a velocity that is \(90 . \%\) of the speed of light. If the wavelength of the particle is \(1.5 \times 10^{-15} \mathrm{m},\) what is the mass of the particle?

Step-by-Step Solution

Answer

The mass of the particle is approximately \(4.283 \times 10^{-28} \mathrm{kg}\).

Step 1: Calculate the velocity of the particle

Since the particle's velocity is 90% of the speed of light, let's calculate the actual velocity: \[v = 0.90c\] where \(c\) is the speed of light, which is approximately \(3.0 \times 10^8 \mathrm{m/s}\).

Step 2: Rewrite the de Broglie wavelength formula using the relativistic momentum formula

Substitute the expression for relativistic momentum, \(p = \frac{m_0v}{\sqrt{1-\frac{v^2}{c^2}}}\), into the de Broglie wavelength formula \(\lambda = \frac{h}{p}\): \[ \lambda = \frac{h}{\frac{m_0v}{\sqrt{1-\frac{v^2}{c^2}}}} \]

Step 3: Solve for the mass of the particle

We are given the wavelength, \(\lambda = 1.5 \times 10^{-15} \mathrm{m}\), and the Planck's constant, \(h \approx 6.626 \times 10^{-34} \mathrm{Js}\). We will now solve for the rest mass, \(m_0\), by rearranging the formula: \[ m_0 = \frac{h\sqrt{1-\frac{v^2}{c^2}}}{v\lambda} \] Now, substitute the given values into the formula: \[ m_0 = \frac{(6.626 \times 10^{-34} \mathrm{Js})\sqrt{1-\frac{(0.90 \times 3.0 \times 10^8 \mathrm{m/s})^2}{(3.0 \times 10^8 \mathrm{m/s})^2}}}{(0.90 \times 3.0 \times 10^8 \mathrm{m/s})(1.5 \times 10^{-15} \mathrm{m})} \]

Step 4: Calculate the mass of the particle

Perform the calculation: \[ m_0 \approx 4.283 \times 10^{-28} \mathrm{kg} \] Thus, the mass of the particle is approximately \(4.283 \times 10^{-28} \mathrm{kg}\).

Key Concepts

de Broglie Wavelength

The concept of de Broglie wavelength bridges the worlds of quantum mechanics and classical physics. It describes the wave-like nature of all particles, not just photons (particles of light). When the famous physicist Louis de Broglie proposed this idea, he suggested that every moving particle or object has an associated wave.

The formula for the de Broglie wavelength (\(\lambda\)) is given by:\[\lambda = \frac{h}{p}\]Where:

• \(\lambda\) is the de Broglie wavelength, measured in meters.
• \(h\) is Planck's constant, around \(6.626 \times 10^{-34}\) Joules per second.
• \(p\) is the momentum of the particle, in kilogram meters per second.

This equation illustrates that the wavelength is inversely proportional to the momentum of the particle. A particle with large momentum will have a smaller wavelength, and vice versa.

In a relativistic context, like when a particle is moving near the speed of light, the momentum involves the rest mass \(m_0\) and velocity \(v\) of the particle. The relativistic momentum \(p\) is given by the expression:\[p = \frac{m_0v}{\sqrt{1-\frac{v^2}{c^2}}} \]This accounts for the fact that as particles approach the speed of light, their mass effectively increases, and thus influences their momentum and de Broglie wavelength.

Planck's Constant

Planck's constant \(h\) is a fundamental quantity in physics that is central to the functionality of quantum mechanics. Named after Max Planck, who originated quantum theory, this constant emerges in numerous fundamental physics equations.

The numerical value of Planck's constant is approximately \(6.626 \times 10^{-34}\) Joules per second. Its role is crucial in calculations dealing with energy and frequency, such as the energy of a photon, given by:\[E = hu \]Where:

• \(E\) is energy (Joules).
• \(u\) is the frequency of the photon (Hertz).

In the realm of matter particles, Planck's constant aids in understanding the wave-particle duality. It is integral to the de Broglie wavelength equation \(\lambda = \frac{h}{p}\), which determines a particle's wavelength as described in the previous section.

Overall, Planck's constant is a bridge connecting the macroscopic and quantum worlds. It offers insights into processes and interactions at the smallest scales that shape our universe.

Rest Mass

Rest mass, often denoted as \(m_0\), is a fundamental concept in both classical and modern physics. It refers to the intrinsic mass of a particle when it is at rest — in other words, not in motion relative to an observer.

In the scenario of particles moving at significant fractions of the speed of light, rest mass becomes part of the calculations for relativistic mass and energy. The rest mass remains constant regardless of the state of motion of the particle.

When dealing with high-speed particles, Einstein's theory of relativity kicks in, introducing the equation:\[E = mc^2 \]which relates mass and energy, with \(c\) representing the speed of light. \(m\) in this context represents relativistic mass.Another way to view rest mass is through its relationship with the momentum of particles. For relativistic particles, the momentum equation integrates rest mass and evaluates their momentum when velocities near the speed of light come into play:\[p = \frac{m_0v}{\sqrt{1-\frac{v^2}{c^2}}} \]Where:

• \(v\) is the velocity of the particle.
• \(c\) is the speed of light.

Thus, rest mass is a central factor in calculating and understanding the behavior of particles at all speeds, playing a key role in both non-relativistic and relativistic physics.