Problem 50 - Physics for Scientists and Engineers with Modern Physics

Question

An electron and a proton, each initially at rest, are accelerated across the same voltage. Assuming that the uncertainty in their position is given by their de Broglie wavelength. find the ratio of the uncertainty in their momentum.

Step-by-Step Solution

Answer

The ratio of uncertainty in momentum is approximately 0.0233.

Step 1: Understanding the Problem

We need to find the ratio of the uncertainty in momentum for an electron and a proton accelerated across the same voltage. Both particles start at rest and their position's uncertainty is given by their de Broglie wavelength. This means connecting potential energy and kinetic energy to de Broglie wavelength.

Step 2: Establishing the Energy Equation

Since both particles are accelerated through the same voltage, they gain the same energy. The kinetic energy gained by a charge accelerated through a voltage \( V \) is given by \( KE = qV \), where \( q \) is the charge of the particle. Both the electron and proton have the same magnitude of charge.

Step 3: Using de Broglie Wavelength Formula

The de Broglie wavelength \( \lambda \) of a particle can be expressed as \( \lambda = \frac{h}{p} \), where \( h \) is Planck's constant and \( p \) is momentum. The momentum uncertainty \( \Delta p \) is related to the uncertainty in position \( \Delta x \) by Heisenberg’s uncertainty principle \( \Delta x \Delta p \approx h \). Here, \( \Delta x = \lambda \).

Step 4: Deriving the Relationship for Each Particle

From de Broglie's equation, we have \( \lambda_e = \frac{h}{p_e} \) and \( \lambda_p = \frac{h}{p_p} \). Using the uncertainty relation \( \Delta x \Delta p \approx h \), and \( \Delta x = \lambda \), then \( \Delta p_e = \frac{h}{\lambda_e} \) and \( \Delta p_p = \frac{h}{\lambda_p} \). Therefore, \( \Delta p_e = p_e \) and \( \Delta p_p = p_p \).

Step 5: Calculate the Ratio of Momentum Uncertainty

\( \Delta p_e / \Delta p_p = p_e / p_p \). From kinetic energy equation \( \frac{1}{2} m v^2 = qV \), solving for \( p = mv \) gives \( p = \sqrt{2m qV} \). So, the momentum for electron \( p_e \) is \( \sqrt{2m_e qV} \) and for proton \( p_p = \sqrt{2m_p qV} \). Their ratio is thus \( \frac{\sqrt{m_e}}{\sqrt{m_p}} \).

Step 6: Final Calculation

The masses of the electron \( m_e \) and proton \( m_p \) are known. Hence, \( \Delta p_e / \Delta p_p = \sqrt{\frac{m_e}{m_p}} \approx \sqrt{\frac{9.109 \times 10^{-31} \, \text{kg}}{1.673 \times 10^{-27} \, \text{kg}}} = \sqrt{5.45 \times 10^{-4}} \). Calculating this, we find the ratio is approximately 0.0233.

Key Concepts

Heisenberg's Uncertainty Principle

Heisenberg's uncertainty principle is a fundamental concept in quantum mechanics. It asserts that there is a limit to how precisely we can simultaneously know certain pairs of physical properties of a particle.
In the context of particle physics, this is often discussed in terms of position and momentum. Simply put, the more precisely we determine the position of a particle, the less precisely we can know its momentum, and vice versa.
This is mathematically expressed as:

• \( \Delta x \Delta p \geq \frac{\hbar}{2} \)

In this equation, \( \Delta x \) represents the uncertainty in position, \( \Delta p \) is the uncertainty in momentum, and \( \hbar \) is the reduced Planck's constant.
The profound implication of this principle is that absolute certainty in measuring both a particle's position and momentum is impossible. This concept lays the groundwork for understanding quantum behavior and highlights the fundamental limits of precision in measurements.

Electron Acceleration

In this context, we consider an electron initially at rest and accelerated by a given voltage. When a charged particle like an electron is accelerated through an electric field, it experiences a gain in kinetic energy.
The kinetic energy (\( KE \)) gained by the electron after being accelerated by a voltage (\( V \)) can be calculated using the formula:

• \( KE = eV \)

Here, \( e \) represents the charge of the electron.
Since the electron was at rest initially, this gained kinetic energy results in the electron's movement, which can be expressed as:

• \( \frac{1}{2}mv^2 = eV \)

By rearranging and solving for velocity (\( v \)), we can relate the electron's velocity to the potential difference.
Once we know the velocity, we can determine the momentum (\( p \)) of the electron using:

• \( p = mv \)

This relationship is crucial for understanding the electron's behavior when its momentum uncertainty is assessed.

Proton Acceleration

Similar to electron acceleration, a proton can also be accelerated by an electric field.
A proton starting at rest will gain kinetic energy when it moves through a potential difference (\( V \)). This energy gain is mirrored in its change in motion.
The kinetic energy acquired by a proton is described as:

• \( KE = eV \)

This expression is similar to that of the electron, given the same voltage and the charge equivalence of protons and electrons.
However, an important distinction lies in the mass of the proton. Protons are considerably heavier than electrons; hence, this influences their velocity under the same conditions:

• \( \frac{1}{2}m_pv^2 = eV \)

Solving for velocity and then momentum through:

• \( p = m_pv \)

We recognize that despite being subjected to the same voltage, a proton's larger mass results in different momentum characteristics compared to an electron.

Momentum Uncertainty

Momentum uncertainty is a direct consequence of Heisenberg's uncertainty principle.
In quantum mechanics, momentum uncertainty (\( \Delta p \)) is inversely related to the accuracy with which we can determine a particle's position (\( \Delta x \)).
This principle plays a pivotal role when determining the uncertainties for particles, like electrons and protons, when they are accelerated. Given the relation:

• \( \Delta x \Delta p \approx h \)

and knowing that their positions measure close to their de Broglie wavelengths, we derive:

• \( \Delta p = \frac{h}{\lambda} \)

With electrons and protons accelerated through the same voltage, their kinetic properties and subsequently momenta impact the calculated momentum uncertainty.
The momentum of a particle is determined by its mass and velocity, leading to a difference in momentum uncertainty due to the differing masses of electrons and protons. This results in distinct ratios of momentum uncertainty when comparing particles under identical acceleration conditions.