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An electron that is accelerated from rest through a potential difference \(V_{0}\) has a de Broglie wavelength \(\lambda_{0}\). What potential difference will double the electron's wavelength? (Express your answer in terms of \(V_{0-}\) )

Short Answer

Expert verified
The new potential difference is \( \frac{V_0}{4} \).

Step by step solution

01

Understand de Broglie Wavelength Formula

The de Broglie wavelength \( \lambda \) of a particle is given by \( \lambda = \frac{h}{p} \), where \( h \) is the Planck's constant and \( p \) is the momentum of the particle. For an electron accelerated through a potential \( V \), its momentum \( p \) can be found using the equation \( p = \sqrt{2m_e e V} \) where \( m_e \) is the electron mass and \( e \) is the elementary charge.
02

Establish Relationship for Initial Condition

Given \( \lambda_0 \) for potential difference \( V_0 \), we have:\[ \lambda_0 = \frac{h}{\sqrt{2m_e e V_0}} \]
03

Express New Wavelength in Terms of Known Quantities

The problem states that the wavelength is doubled, so the new wavelength \( \lambda = 2\lambda_0 \). Substitute into the de Broglie formula:\[ 2\lambda_0 = \frac{h}{\sqrt{2m_e e V}} \]
04

Draw Parallel Between New and Original Conditions

Set up the equation for the doubled wavelength with respect to \( V \) and original conditions using two equations:1. \( \lambda_0 = \frac{h}{\sqrt{2m_e e V_0}} \)2. \( 2\lambda_0 = \frac{h}{\sqrt{2m_e e V}} \)Using the relation, we find:\[ \frac{h}{2\lambda_0} = \frac{h}{\lambda_0}\cdot \frac{1}{2} = \frac{\sqrt{2m_e e V_0}}{2} \]
05

Solve for New Potential Difference

Square both sides of the equation from step 4 to solve for \( V \):\[ 2m_e e V = \left(\frac{\sqrt{2m_e e V_0}}{2}\right)^2\]Solving yields:\[ V = \frac{V_0}{4} \]
06

Conclusion

If the wavelength is to be doubled, then the potential difference \( V \) in terms of the initial potential difference \( V_0 \) is \( \frac{V_0}{4} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electron Acceleration
Electrons are fascinating little particles. When they hover around in their natural state, they have a certain amount of energy. However, when we apply a potential difference or voltage across them, something exciting happens: they accelerate! This acceleration means the electron picks up speed and energy, shifting from a resting state to a dynamic one.
  • An electron starts from rest, with very low velocity.
  • Applying a potential difference makes the electron gain kinetic energy.
  • The kinetic energy gained can be calculated using the equation: \( KE = eV \), where \( e \) is the charge of the electron and \( V \) is the potential difference.

As the electron speeds up, its momentum increases, and this change is crucial for determining its behavior quantum mechanically, such as calculating its de Broglie wavelength. Increased momentum effectively shrinks the wavelength, illustrating the wave-particle duality of the electron. This behavior is key to understanding electron motion in electric fields.
Potential Difference
The concept of "potential difference" is fundamental in classical and quantum physics as it deals with the difference in voltage between two points. By applying this electrical potential across an electron, we cause it to accelerate, gaining kinetic energy from the electric field.
  • Potential difference, measured in volts, refers to the energy difference or electrical tension that pushes electrons through a conductor.
  • The equation used is \( V = \frac{W}{Q} \), where \( W \) is the work done and \( Q \) is the charge of the electrons.
  • In our exercise: a potential difference \( V_0 \) accelerates an electron to move through a space, altering its de Broglie wavelength.

Its essential role lies in controlling electron speed, and consequently, their properties and behaviors. By adjusting the potential difference in our calculations, we alter the outcome on electron movement and their matching de Broglie wavelengths.
Planck's Constant
Planck's constant is an essential piece in the puzzle of quantum mechanics, acting as a bridge between the particle and wave nature of matter. It's symbolized by \( h \) and has a very small value, but its effect is profound. In equations relating to the de Broglie wavelength, Planck's constant helps decipher the wave properties of particles, including electrons.
  • Planck's constant is approximately \( 6.626 \times 10^{-34} \text{ Js} \).
  • In the de Broglie equation \( \lambda = \frac{h}{p} \), it connects the wavelength \( \lambda \) of a particle to its momentum \( p \).
  • Thus, it plays a critical role in understanding phenomena like wave-particle duality.

This constant is vital in explaining why electrons, despite being particles, exhibit wave-like characteristics under certain conditions. By multiplying a particle's momentum, Planck's constant can ripple through realms of classical physics, allowing the exploration of deeper quantum behaviors. This duality becomes evident in practical applications like the de Broglie wavelength of an accelerated electron, as seen when it is subjected to different potential differences.

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