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In the early 20th century, scientists discovered that light has both wave-like and particle-like properties. This was a major breakthrough that led to the concept of wave-particle duality. On one hand, phenomena like interference and diffraction clearly showed light behaving as a wave. On the other hand, experiments such as the photoelectric effect demonstrated that light can also behave as a particle, called a photon.
This dual nature challenged the classical picture, which treated waves and particles as fundamentally different. The natural question arose: could this wave-particle duality also apply to matter particles like electrons and protons?
Answering this was the key to understanding many mysteries in physics and led to the foundation of modern quantum mechanics.
The French physicist Louis de Broglie proposed a revolutionary idea in 1924: every particle of matter has a wave associated with it. Just as light can behave as a particle and a wave, particles like electrons also have wave-like properties. This idea is called the De Broglie hypothesis.
How do we express this concept quantitatively? De Broglie connected the wavelength \( \lambda \) of the matter wave to the momentum \( p \) of the particle through a simple formula:
Recall that momentum \( p \) for a particle with mass \( m \) and velocity \( v \) is given by:
This diagram shows a particle moving with momentum \( p \) and an associated wave with wavelength \( \lambda \). Notice the inverse relation: as momentum increases, wavelength decreases.
The De Broglie hypothesis means that matter, which we usually think of as discrete particles, actually has wave characteristics. This is significant especially for very small particles like electrons, where the wavelength can be comparable to atomic scales. For large objects like a cricket ball, the wavelength is so tiny it is practically undetectable.
This dual nature laid down the foundation for understanding atomic and subatomic phenomena, such as electron diffraction and quantum mechanics.
The De Broglie hypothesis was a bold theory, but how do we test if particles really have wave nature? The answer came from the Davisson-Germer experiment in 1927, which showed electron diffraction-a characteristic wave behavior.
In this experiment, a beam of electrons was directed at a crystal of nickel atoms. The crystal acted like a diffraction grating for electrons. If electrons behaved purely like particles, we would expect a simple scattering pattern. Instead, an interference pattern was observed, similar to waves diffracting through a slit.
graph TD A[Electron Gun] --> B[Beam of electrons] B --> C[Incident on Nickel Crystal] C --> D[Diffraction of electrons] D --> E[Detection of peaks at specific angles] E --> F[Confirm De Broglie wavelength]
The angles and intensities of the diffraction peaks matched exactly those predicted by the De Broglie wavelength calculated for the electrons' momentum. This experiment confirmed the wave nature of matter.
Electron diffraction later became a crucial tool for studying crystal structures, and the wave-particle duality principle became a cornerstone of modern physics.
Step 1: Calculate the momentum \(p = mv\).
\( p = 9.11 \times 10^{-31} \times 1.5 \times 10^{6} = 1.3665 \times 10^{-24}\, \mathrm{kg\cdot m/s} \)
Step 2: Use the De Broglie formula:
\( \lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34}}{1.3665 \times 10^{-24}} = 4.85 \times 10^{-10}\, \mathrm{m} \)
Answer: The wavelength of the electron is approximately \(0.485\, \mathrm{nm}\), which is of atomic scale.
Electron:
Mass \( m_e = 9.11 \times 10^{-31} \, \mathrm{kg} \), Velocity \( v_e = 1 \times 10^{6} \, \mathrm{m/s} \)
Momentum \( p_e = m_e v_e = 9.11 \times 10^{-31} \times 10^{6} = 9.11 \times 10^{-25} \, \mathrm{kg\cdot m/s} \)
Wavelength \( \lambda_e = \frac{6.626 \times 10^{-34}}{9.11 \times 10^{-25}} = 7.27 \times 10^{-10} \, \mathrm{m} \)
Cricket ball:
Mass \( m_c = 0.16\, \mathrm{kg} \), Velocity \( v_c = 40\, \mathrm{m/s} \)
Momentum \( p_c = 0.16 \times 40 = 6.4\, \mathrm{kg\cdot m/s} \)
Wavelength \( \lambda_c = \frac{6.626 \times 10^{-34}}{6.4} = 1.04 \times 10^{-34}\, \mathrm{m} \)
Explanation: The electron's wavelength is about one-tenth of a nanometer (atomic scale), which can cause observable diffraction effects. The cricket ball's wavelength is unimaginably small, far less than the size of an atomic nucleus, making its wave nature undetectable in everyday life.
Step 1: Use the de Broglie relation: \( p = \frac{h}{\lambda} \).
Given \( \lambda = 0.05 \times 10^{-9} = 5 \times 10^{-11} \, \mathrm{m} \).
Calculate momentum:
\( p = \frac{6.626 \times 10^{-34}}{5 \times 10^{-11}} = 1.325 \times 10^{-23} \, \mathrm{kg\cdot m/s} \)
Answer: The electron momentum is \(1.325 \times 10^{-23}\, \mathrm{kg\cdot m/s}\).
Step 1: Find velocity of electrons using the kinetic energy from acceleration:
Kinetic energy \( E_k = eV = 1.6 \times 10^{-19} \times 150 = 2.4 \times 10^{-17} \, \mathrm{J} \)
Electron mass \( m = 9.11 \times 10^{-31} \, \mathrm{kg} \)
Velocity \( v = \sqrt{\frac{2E_k}{m}} = \sqrt{\frac{2 \times 2.4 \times 10^{-17}}{9.11 \times 10^{-31}}} \approx 7.26 \times 10^{6} \, \mathrm{m/s} \)
Step 2: Calculate de Broglie wavelength:
Momentum \( p = mv = 9.11 \times 10^{-31} \times 7.26 \times 10^{6} = 6.615 \times 10^{-24} \, \mathrm{kg\cdot m/s} \)
\( \lambda = \frac{6.626 \times 10^{-34}}{6.615 \times 10^{-24}} = 1.001 \times 10^{-10} = 0.1001\, \mathrm{nm} \)
Step 3: Use Bragg's law for diffraction maxima:
\( 2d \sin \theta = n \lambda \) where \( n=1 \) (first order), \( d = 0.25\, \mathrm{nm} \)
\( 2 \times 0.25 \sin \theta = 0.1001 \implies \sin \theta = \frac{0.1001}{0.5} = 0.2002 \)
\( \theta = \sin^{-1}(0.2002) \approx 11.54^\circ \)
Answer: The first-order diffraction maximum occurs at an angle of approximately \(11.54^\circ\).
Step 1: Convert energies to joules:
\(E_1 = 100 \times 1.6 \times 10^{-19} = 1.6 \times 10^{-17}\, \mathrm{J}\)
\(E_2 = 400 \times 1.6 \times 10^{-19} = 6.4 \times 10^{-17}\, \mathrm{J}\)
Step 2: Calculate wavelengths using:
\( \lambda = \frac{h}{\sqrt{2mE_k}} \) (since \(p = \sqrt{2mE_k}\))
For \(E_1\):
\( \lambda_1 = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 9.11 \times 10^{-31} \times 1.6 \times 10^{-17}}} \approx 1.23 \times 10^{-10} \, \mathrm{m} \)
For \(E_2\):
\( \lambda_2 = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 9.11 \times 10^{-31} \times 6.4 \times 10^{-17}}} \approx 6.14 \times 10^{-11} \, \mathrm{m} \)
Step 3: Calculate change in wavelength:
\( \Delta \lambda = \lambda_1 - \lambda_2 = (1.23 - 0.614) \times 10^{-10} = 6.16 \times 10^{-11} \, \mathrm{m} \)
Answer: Increasing the kinetic energy fourfold decreases the electron's wavelength by approximately \(6.16 \times 10^{-11} \, \mathrm{m}\), or halves it.
When to use: When kinetic energy is provided in eV during calculations involving velocity or momentum.
When to use: To speed up numerical calculations while maintaining reasonable precision.
When to use: Before solving any numerical problem to avoid calculation errors.
When to use: When dealing with electron diffraction or crystal lattice problems.
When to use: To quickly judge feasibility of wave behavior in given physics problems.
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