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Magazine Chapter 28: Problem 44
\(\bullet\) (a) What is the de Broglie wavelength of an electron accelerated through 800 \(\mathrm{V} ?\) (b) What is the de Broglie wavelength of a proton accelerated through the same potential difference?
Short Answer
Expert verified
Electron's wavelength is much smaller than proton's due to the mass difference.
Step by step solution
01
Understand the de Broglie Wavelength Formula
The de Broglie wavelength \( \lambda \) of a particle is given by\[\lambda = \frac{h}{p}\]where \( h \) is the Planck constant \( 6.626 \times 10^{-34} \mathrm{~m}^2 \mathrm{~kg/s} \) and \( p \) is the momentum of the particle. For a particle with mass \( m \) and velocity \( v \), the momentum \( p \) is given by \( p = mv \). For a charged particle accelerated through a potential difference \( V \), the velocity \( v \) can be found using energy equations.
02
Calculate Electron's Velocity
For an electron accelerated through 800 V, the kinetic energy gained is equal to the potential energy, \( qV \), where \( q \) is the charge of an electron \( 1.602 \times 10^{-19} \mathrm{~C} \). Using the equation:\[qV = \frac{1}{2}mv^2\]Substitute for \( v \) to find velocity:\[v = \sqrt{ \frac{2qV}{m} }\]where the mass of an electron \( m_e = 9.109 \times 10^{-31} \mathrm{~kg} \). Substitute the values to find \( v \).
03
Calculate De Broglie Wavelength for Electron
Using the velocity found in Step 2, calculate the momentum \( p = m_ev \). Then substitute into the de Broglie wavelength formula:\[\lambda = \frac{h}{m_ev}\]Substitute the calculated velocity and constants to find \( \lambda \) for the electron.
04
Calculate Proton's Velocity
For a proton accelerated through the same potential difference, use the same energy equation:\[qV = \frac{1}{2}mv^2\]But here, use the mass of a proton \( m_p = 1.673 \times 10^{-27} \mathrm{~kg} \). Calculate \( v \) using the same formula by substituting the mass of the proton.
05
Calculate De Broglie Wavelength for Proton
Calculate the proton's momentum \( p = m_pv \) using the velocity found for the proton. Then, substitute into the de Broglie wavelength formula:\[\lambda = \frac{h}{m_pv}\]Substitute the calculated velocity and constants to get \( \lambda \) for the proton.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Planck constant
The Planck constant, denoted by the letter \( h \), is a fundamental constant in quantum mechanics. It has a value of \( 6.626 \times 10^{-34} \, \text{m}^2 \text{kg/s} \) and is central to the calculation of the de Broglie wavelength.
This constant represents the smallest action that can occur in nature and ties together the concepts of energy and frequency.
e.g.,
This constant represents the smallest action that can occur in nature and ties together the concepts of energy and frequency.
e.g.,
- The formula \( E = hf \) uses the Planck constant to relate energy \( E \) and frequency \( f \).
- In the context of de Broglie wavelength, it helps relate the wavelength to momentum through \( \lambda = \frac{h}{p} \).
Momentum
Momentum \( p \) is a measure of the motion of an object and is a key component when calculating the de Broglie wavelength.
For a particle moving with velocity \( v \) and having mass \( m \), momentum is calculated as \( p = mv \).
This is what connects motion to wave properties through the de Broglie equation:
For a particle moving with velocity \( v \) and having mass \( m \), momentum is calculated as \( p = mv \).
This is what connects motion to wave properties through the de Broglie equation:
- The de Broglie wavelength is found by \( \lambda = \frac{h}{p} \).
- Momentum represents how much motion energy a particle has, influencing its wavelength.
Electron acceleration
When an electron is accelerated through an electric potential, its energy increases. This energy boost is often represented as \( qV \), where \( q \) is the charge of the electron (\( 1.602 \times 10^{-19} \, \text{C} \)) and \( V \) is the potential difference.
For an electron accelerated through a voltage, the kinetic energy gained equals the electrical energy given, or:\[ qV = \frac{1}{2} mv^2 \]This equation helps us find the electron's velocity after acceleration: \( v = \sqrt{ \frac{2qV}{m_e} } \). The mass of the electron \( m_e \) is \( 9.109 \times 10^{-31} \, \text{kg} \). This velocity then allows calculation of an electron's momentum for use in the de Broglie equation, showing how voltage conversion translates into particle motion.
For an electron accelerated through a voltage, the kinetic energy gained equals the electrical energy given, or:\[ qV = \frac{1}{2} mv^2 \]
Proton acceleration
Similar to electrons, protons can be accelerated to gain kinetic energy through a potential difference. However, because protons are much heavier, with a mass \( m_p \) of \( 1.673 \times 10^{-27} \, \text{kg} \), they respond differently than electrons.
The energy relationship is again \( qV = \frac{1}{2} mv^2 \). To find the proton's velocity after being accelerated through the same potential \( V \): \[ v = \sqrt{ \frac{2qV}{m_p} } \]
This velocity, significantly lower due to the larger mass, results in a different momentum \( p = m_pv \). The de Broglie wavelength calculations will show longer wavelengths compared to the electron. By examining these equations, one discovers distinct behaviors in particles based on their mass, in response to the same energy input.
The energy relationship is again \( qV = \frac{1}{2} mv^2 \). To find the proton's velocity after being accelerated through the same potential \( V \): \[ v = \sqrt{ \frac{2qV}{m_p} } \]
