Problem 6 - College Physics

Question

Electrons are accelerated from rest through an electric potential difference. (a) If this potential difference increases to four times the original value, the new de Broglie wavelength will be (1) four times, (2) twice, (3) one-fourth, (4) one-half that of the original. Why? (b) If the original potential is \(250 \mathrm{kV}\) and the new potential is \(600 \mathrm{kV}\), what is the ratio of the new de Broglie wavelength to the original?

Step-by-Step Solution

Answer

(a) The new wavelength is one-half the original. (b) The ratio is \(0.645\).

Step 1: Understand the de Broglie Wavelength Formula

The de Broglie wavelength \( \lambda \) of a particle is given by the formula \( \lambda = \frac{h}{p} \), where \( h \) is Planck's constant and \( p \) is the momentum of the particle. For an electron accelerated through a potential difference \( V \), the kinetic energy gained is equal to the work done by the electric field, which is \( eV \).

Step 2: Relate Kinetic Energy to Momentum

The kinetic energy \( K.E. \) is \( eV \) and it is also given by \( \frac{1}{2}mv^2 \). The momentum \( p \) of an electron is related to its kinetic energy by \( p = \sqrt{2mK.E.} = \sqrt{2meV} \). As such, the de Broglie wavelength becomes \( \lambda = \frac{h}{\sqrt{2meV}} \).

Step 3: Analyze the Effect of Increasing Potential

If the potential difference is increased to four times its initial value, \( V \to 4V \), the new momentum becomes \( p = \sqrt{2m(4eV)} = 2\sqrt{2meV} \). Thus, the new wavelength \( \lambda' = \frac{h}{2\sqrt{2meV}} = \frac{\lambda}{2} \) because \( \sqrt{4V} = 2\sqrt{V} \). So, the new wavelength is half of the original.

Step 4: Calculate the Ratio for Given Potentials

Given \( 250 \mathrm{kV} \) as the original potential and \( 600 \mathrm{kV} \) as the new potential, the ratio of the new wavelength \( \lambda' \) to the original wavelength \( \lambda \) is \( \frac{\lambda'}{\lambda} = \frac{\sqrt{V}}{\sqrt{V'}} = \sqrt{\frac{250}{600}} \).

Step 5: Computation of Ratio

Calculate \( \sqrt{\frac{250}{600}} \). First simplify the ratio: \( \frac{250}{600} = \frac{5}{12} \). Then find the square root of this ratio: \( \sqrt{\frac{5}{12}} \approx 0.645 \).

Step 6: Conclusion

The new de Broglie wavelength is approximately 64.5% of the original wavelength when the potential is increased from 250 kV to 600 kV.

Key Concepts

Electron Acceleration

When we talk about electron acceleration, we often refer to the process where electrons gain speed as they move through an electric field. This is important in understanding many physical phenomena and technologies, such as cathode ray tubes and particle accelerators.
In the context of our exercise, electrons start from rest and are accelerated by an electric potential difference. As they accelerate, they gain kinetic energy, which is directly related to their velocity. This increase in velocity then impacts their de Broglie wavelength, the focus of our study.
Electrons, being charged particles, respond strongly to electric fields. When an electric potential difference is applied, it creates an electric field that performs work on the electrons. The work done on the electrons by this field is responsible for their acceleration.
To simplify, think of electrons as tiny balls that roll faster down a hill as it becomes steeper—except in this case, the 'steepness' is the electric potential difference.

Electric Potential Difference

Electric potential difference, often referred to simply as voltage, is the work needed to move a unit charge from one point to another. This concept is fundamental to understanding how circuits operate and how electrons are set into motion within a conductor.

• The potential difference is measured in volts (V), where one volt equates to the energy (in joules) per coulomb of charge.
• In our exercise, the electrons are accelerated by this electric potential, which essentially 'pushes' them to gain speed and kinetic energy.
• It is this potential difference that determines how much kinetic energy the electrons will gain. If you increase the voltage, you increase the energy imparted to the electrons, thus increasing their speed and, consequentially, affecting their de Broglie wavelength.

The relationship between potential difference and energy is a simple yet powerful one, linking the concepts of electrical work and motion in elegantly unified physics principles.

Kinetic Energy

Kinetic energy is the energy that an object possesses due to its motion. For electrons accelerated from rest, their kinetic energy comes entirely from the work done by the electric field.

• The formula for kinetic energy is given by: \[ K.E. = \frac{1}{2}mv^2 \]where \( m \) is the mass of the electron, and \( v \) is its velocity.
• However, when we relate this to electric potential difference, the kinetic energy can also be expressed as \( eV \) (where \( e \) is the charge of the electron, and \( V \) is the potential difference).
• This shows that the kinetic energy gained by an electron is directly proportional to the potential difference it goes through. Doubling the voltage will double the kinetic energy acquired by the electron, allowing it to move faster and thus change its momentum and de Broglie wavelength accordingly.

Kinetic energy not only plays a vital role in the context of individual particles but also in the broader understanding of energy conversion and conservation in physics.

Momentum

Momentum is a fundamental concept in physics that describes the quantity of motion an object has. It is directly tied to both the mass and velocity of an object, given by the formula:
\[ p = mv \]In the case of electrons, momentum is key to understanding the de Broglie wavelength. When electrons are accelerated through an electric potential difference, their velocity increases, leading to a rise in momentum.

• Momentum for an electron accelerated by a potential \( V \) is given by: \[ p = \sqrt{2meV} \]
• As the potential difference increases, so does the momentum of the electron. This increase reduces the de Broglie wavelength because \[ \lambda = \frac{h}{p} \].A higher momentum results in a smaller wavelength.
• This highlights the inverse relationship between momentum and the de Broglie wavelength, a fundamental aspect for understanding electron behavior at the quantum level.

Momentum connects not only with the kinetic energy of the electron but also with how we perceive wave-particle duality in quantum physics, bridging classical mechanics with quantum mechanics.