Topic 4: Waves and particle Nature of light (Part 3)

Topic 4: Content
86)    Understand how diffraction experiments provide evidence for the wave nature of electrons.
87)    Be able to use the de Broglie equation = [math] \frac{h}{p} [/math]
88)    Understand that waves can be transmitted and reflected at an interface between media.
89)    Understand how a pulse-echo technique can provide information about the position of an object and how the amount of information obtained may be limited by the wavelength of the radiation or by the duration of pulses.
90)    Understand how the behavior of electromagnetic radiation can be described in terms of a wave model and a photon model, and how these models developed over time.
91)    Be able to use the equation [math] E = hf [/math],that relates the photon energy to the wave frequency.
92)    Understand that the absorption of a photon can result in the emission of a photoelectron.
93)    Understand the terms threshold frequency and work function and be able to use the equation [math] hf = \phi + \frac{1}{2} mv_{\text{max}}^2 [/math]
94)    Be able to use the electron volt (eV) to express small energies.
95)    Understand how the photoelectric effect provides evidence for the particle nature of electromagnetic radiation.
96)    Understand atomic line spectra in terms of transitions between discrete energy levels and understand how to calculate the frequency of radiation that could be emitted or absorbed in a transition between energy levels.

1. Electron diffraction:

  • In 1912, Max von Laue suggested that if X-rays had a wavelength similar to atomic separations, they should produce diffraction patterns when fired through single crystal materials.
  • He earned a Nobel Prize for his work, and X-ray diffraction techniques have developed so that they now are able to explore the structures of the most complex molecules.
  • In von Laue’s time, electrons had been shown to be particles of mass [math] 9.1 × 10^{-31}kg [/math] carrying a charge of [math] 1.6 × 10^{-19}C [/math].
  • The idea that particles like electrons could also behave as waves was proposed in 1924 by Louis de Broglie.
  • [math] λ = \frac{h}{mv} [/math]
  • Where h is Planck’s constant ([math] 6.6 × 10^{-34} J.s [/math]). This is often referred to as the ‘de Broglie wavelength’ of the electron
  • This equation links the wavelength to the momentum, (mv), of the ‘particle’.
  • The relationship was subsequently verified using electron diffraction and can be demonstrated using the vacuum tube shown in Figure 1.
  • High-velocity electrons are fired from the electron gun at the graphite crystal.
  • A voltage of 1kV gives the electron a speed of about [math] 2 × 10^7ms^{-1}[/math]  and hence a de Broglie wavelength of around [math] 4 × 10^{-11} m[/math]
  • The thin graphite crystal has a regular hexagonal structure with atomic separations similar to this wavelength.
  • It behaves a bit like a three-dimensional diffraction grating and creates a diffraction pattern of concentric rings on the fluorescent screen.
  • When the voltage across the electron gun is increased, the faster moving electrons have a shorter wavelength so the diffraction is less and the diameter of the rings is reduced.
  • Figure 1 Electron diffraction

2.The De-Broglie equation:

  • Particles can sometimes behave like waves and waves can sometimes behave like particles.
  • These ideas emerged from the theory of quantum mechanics, which was developed in the early twentieth century.
  • In 1924, de Broglie linked waves and particles by the formula:
  • [math] λ = \frac{h}{p} [/math]
  • where λ is the wavelength corresponding to a particle of momentum p (i.e., a particle of mass m travelling with a velocity v).
  • Thus, was born the concept of wave–particle duality.

⇒ Examples:

    1. Calculate the wavelength of a football of mass 440g travelling at [math]30ms^{-1}[/math] and hence explain why it does not show wave-like properties.
    2. a) Show that an electron of energy 15keV has a wavelength of about [math]10^{-11} m[/math].
    3. b) In which region of the electromagnetic spectrum is this wavelength?
  • Solution:
  • (1)Given data:
    Mass of the football = m = 440g = 0.44 kg
    Velocity of moving football = v = 30 m/s
    Planck’s constant = h = [math] 6.63 \times 10^{-34} J.s [/math]
    Wavelength = λ =?
    Formula:
  • [math] \lambda = \frac{h}{p} \\
    \lambda = \frac{h}{mv} [/math]
  • Solution:
  • [math] \lambda = \frac{h}{mv} \\
    \lambda = \frac{6.63 \times 10^{-34}}{(0.44)(30)} \\
    \lambda = 5 \times 10^{-35} \text{ m} [/math]
  • This wavelength is far too short for us to observe any wave properties of the football.
  • (2)
    Given data:
    Kinetic energy of electron [math]= \text{k.E} = 15 \, \text{keV} = 15 \times 10^3 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J} \, \text{eV}^{-1} = 2.4 \times 10^{-15} \, \text{J} [/math]
    Mass of electron [math] m = 9.11 \times 10^{-31} \, \text{kg} [/math]
    Wavelength = λ =?
    Formula:
  • [math] \text{k.E} = \frac{1}{2} mv^2 \\
    v = \sqrt{\frac{2 \cdot \text{k.E}}{m}} \\
    \lambda = \frac{h}{mv} [/math]
  • Solution:
    1.  [math]v = \sqrt{\frac{2 \cdot (2.4 \times 10^{-15})}{(9.11 \times 10^{-31})}} \\
      v = 7.26 \times 10^7 \, \text{m/s} \\
      \lambda = \frac{h}{mv} \\
      \lambda = \frac{6.63 \times 10^{-34}}{(9.11 \times 10^{-31})(7.26 \times 10^7)} \\
      \lambda = 1 \times 10^{-11} \, \text{m} [/math]
    2. This wavelength is in the X-ray region of the electromagnetic spectrum.

3. Waves can be transmitted and reflected at an interface between media.

  • The transmission and reflection of waves at an interface between media is a fundamental concept. Here are some key points to note:
  • Transmission:
    – When a wave reaches an interface, it can pass through and continue propagating in the new medium.
    – The transmitted wave has the same frequency but may have a different speed, wavelength, and amplitude.
  • Figure 2 Waves continue their propagation in the new medium
  • Reflection:
    – When a wave reaches an interface, it can bounce back into the original medium.
    – The reflected wave has the same frequency and speed as the incident wave but may have a different amplitude and phase.
  • Figure 3 Waves bounce back into the original medium
  • Factors affecting transmission and reflection:
    – Acoustic impedance: The ratio of pressure to flow velocity in a medium.
    – Angle of incidence: The angle at which the wave approaches the interface.
    – Wavelength: The distance between successive wavefronts.
    – Frequency: The number of oscillations per second.
  • Types of reflection:
    – Specular reflection: Mirrors-like reflection, where the angle of reflection equals the angle of incidence.
    – Diffuse reflection: Scattered reflection, where the wave is dispersed in various directions.
  • Applications:
    – Medical imaging: Ultrasound waves transmitted and reflected in the body help create images.
    – Seismic exploration: Waves transmitted and reflected in the Earth’s interior help locate subsurface structures.
    – Optical fibers: Light waves transmitted through fibers with minimal reflection enable data transmission.
  • Remember, understanding wave transmission and reflection is crucial in physics, engineering, and many other fields.

4. Pulse-echo technique:

  • The pulse-echo technique is a method used to determine the position of an object by emitting a pulse of radiation (e.g., sound wave, electromagnetic wave) and measuring the time delay between the emitted pulse and the reflected pulse that bounces back from the object.
  • Here’s how it works:
    1. Emit a pulse of radiation towards the object.
    2. Measure the time delay (t) between the emitted pulse and the reflected pulse.
    3. Calculate the distance (d) of the object using the formula: [math] d = c \times \frac{t}{2} [/math]
      , where c is the speed of the radiation.
  • The amount of information obtained may be limited by:
    1. Wavelength: If the wavelength is too long, the resolution will be poor, making it difficult to accurately determine the object’s position.
    2. Pulse duration: If the pulse duration is too long, the pulse will overlap with the reflected pulse, making it difficult to measure the time delay accurately.
  • Shorter wavelengths and shorter pulse durations provide higher resolution and more accurate positioning.

5. The behavior of electromagnetic radiation in terms of a wave model and a photon model:

  • Let’s dive into the wave model and photon model of electromagnetic radiation.
  • Wave Model:
  • Describes electromagnetic radiation as a wave propagating through space
  • Figure 4 electromagnetic radiation as a wave propagating
  • Characterized by:
    – Wavelength (λ)
    – Frequency (f)
    – Speed (c)
    – Electric and magnetic field components
  • Developments:
    – James Clerk Maxwell (1864): Formulated equations describing electromagnetic waves
    – Heinrich Hertz (1887): Experimentally confirmed the existence of electromagnetic waves
  • Photon Model:
  • Describes electromagnetic radiation as particles (photons) with energy and momentum
  • Figure 5 electromagnetic radiation propagation as particles
  • Characterized by:
    – Energy (E)
    – Momentum (p)
    – Frequency (f)
    – Wavelength (λ)
  • Developments:
    – Albert Einstein (1905): Introduced the concept of light quanta (photons)
    – Arthur Compton (1923): Experimentally confirmed the particle-like behavior of photons
  • Evolution of Models:
    1. Classical Wave Model (1800s): Electromagnetic radiation was thought to be a wave in a hypothetical medium called the “ether”.
    2. Maxwell’s Equations (1864): Unified electricity and magnetism, predicting electromagnetic waves.
    3. Photon Model (1905): Einstein introduced light quanta, challenging the classical wave model.
    4. Wave-Particle Duality (1920s): Experiments showed that electromagnetic radiation exhibits both wave-like and particle-like behavior.
    5. Quantum Electrodynamics (1940s): A quantum field theory merging the wave and photon models.
  • Remember, our understanding of electromagnetic radiation has evolved significantly over time, and both the wave and photon models are essential for describing its behavior.

6. The photon energy to wave frequency:

  • Bohr proposed that the electron orbits the proton like a satellite orbits the Earth, with the necessary force being provided by the attractive nature of the electrical charges of opposite sign on the electron (negative) and proton (positive).
  • Just as satellites can orbit at different distances from the Earth, Bohr reasoned that the electron could be given energy (or excited) and therefore exist in less-stable, higher-energy orbits.
  • If sufficient energy is supplied, the electron can escape completely from its atom. This is called ionization, and the energy needed for an electron to escape from its lowest energy level is called the ionization energy of the atom.
  • By applying quantum theory to his model, Bohr concluded that the electron could only exist in certain discrete or quantized orbits.
  • This means orbits having distinct, fixed amounts of energy. As a particular value of energy is associated with each orbit, we say that the electrons have discrete or quantized energy levels.
  • When an electron drops from a higher energy level to a lower level, it gives out the energy difference in the form of one quantum of radiation, hf
  • [math] E = hf [/math]
  • Where:
    – E is the energy of the photon (in joules, J)
    – h is Planck’s constant (approximately [math]6.626 \times 10^{-34} \text{ J.s}[/math])
    – f is the frequency of the photon (in hertz, Hz)
  • This equation states that the energy of a photon is directly proportional to its frequency. The higher the frequency, the higher the energy of the photon.
  • Thus, when an electron drops from energy level [math]E_2 [/math] to a level [math] E_1 [/math]:
  • [math] hf = E_2 – E_1 [/math]

  • Figure 6 electron drops from energy level higher state to lower state
  • and we get an emission spectrum.
  • Examples:
  • Use the data in Figure 7 to answer the following questions.

    Figure 7 Energy diagram

    1. Calculate the energy in joules that would have to be supplied for an electron at the lowest energy level to escape from the atom (i.e., the ionization energy of the atom).
    2. What is the wavelength of the spectral line emitted when an electron falls from the −1.51eV level to the −3.41eV level?
      Suggest what color this line would be.
    3. Between which energy levels must an electron fall to emit a blue line of wavelength 434nm?

    Solution:

  • (1)
  • Ionization energy =  [math] +13.6 \ \text{eV} [/math]
  • [math] = 13.6 \ \text{eV} \times 1.6 \times 10^{-19} \ \text{J eV}^{-1} \\
    = 2.2 \times 10^{-18} \ \text{J} [/math]
  • (2)
  • [math] hf = E_2 – E_1 \\
    = (-1.51 \ \text{eV}) – (-3.41 \ \text{eV}) = 1.90 \ \text{eV} \\
    = 1.90 \ \text{eV} \times 1.6 \times 10^{-19} \ \text{J eV}^{-1} \\
    = 3.04 \times 10^{-19} \ \text{J} \\
    f = \frac{3.04 \times 10^{-19} \ \text{J}}{6.63 \times 10^{-34} \ \text{Js}} \\
    = 4.59 \times 10^{14} \ \text{Hz} \\
    c = f \lambda \\
    \lambda = \frac{c}{f} \\
    = \frac{3.00 \times 10^8 \ \text{ms}^{-1}}{4.59 \times 10^{14} \ \text{s}^{-1}} \\
    = 6.54 \times 10^{-7} \ \text{m} \\
    = 654 \ \text{nm} [/math]
  • This is visible red light.
  • (3)
  • [math] f = \frac{c}{\lambda} \\
    = \frac{3.00 \times 10^8 \ \text{ms}^{-1}}{434 \times 10^{-9} \ \text{m}} \\
    = 6.91 \times 10^{14} \ \text{Hz} \\
    E_2 – E_1 = hf \\
    = 6.63 \times 10^{-34} \ \text{Js} \times 6.91 \times 10^{14} \ \text{s}^{-1} \\
    = 4.58 \times 10^{-19} \ \text{J} \\
    = \frac{4.58 \times 10^{-19} \ \text{J}}{1.6 \times 10^{-19} \ \text{J eV}^{-1}} \\
    = 2.86 \ \text{eV} [/math]
  • This could arise from an electron falling from the −0.54eV level to the −3.41 eV level (allowing for rounding differences).

7. The absorption of a photon-Emission of a photoelectron:

  • The absorption of a photon can indeed result in the emission of a photoelectron. This phenomenon is known as the photoelectric effect.
  • Figure 8 photoelectric effect (electrons emit only at 3.1ev with 400nm wavelength)
  • – Absorption of a photon: A photon collides with an atom or molecule, transferring its energy to the electron.
    – Emission of a photoelectron: The energized electron gains enough energy to escape the atom or molecule, becoming a free electron.
  • The photoelectric effect is a fundamental concept in physics, demonstrating the particle-like behavior of light and the wave-particle duality.
    – The energy of the absorbed photon must be greater than the binding energy of the electron.
    – The emitted photoelectron’s energy is dependent on the frequency, not intensity, of the incident light.
    – The photoelectric effect is a quantum mechanical phenomenon, as it involves the interaction between light and matter at the atomic scale.
  • Applications:
    – Photoelectric cells (figure 8)
    – Solar cells
  • Figure 9 working principle of solar cell
  • – Electron microscopy
  • Figure 10 Function of electron microscope
  • – Spectroscopy
  • Figure 11 Raman spectroscopy

8. Einstein’s photoelectric equation:

  • If a photon of energy hf has more than the bare minimum energy needed to just remove an electron from the surface of a metal (called the work function, symbol Φ), the remaining energy is given to the electron as kinetic energy, [math] \frac{1}{2} mv^2 [/math].
  • Applying the conservation of energy, the maximum kinetic energy [math] \frac{1}{2} mv_{\text{max}}^2 [/math] that an electron can have been given by:
  • [math] hf = \phi + \frac{1}{2} mv_{\text{max}}^2 [/math]
  • This is known as Einstein’s photoelectric equation.
  • Electrons emitted from further inside the metal will need more than the work function to escape and so will have less than this maximum kinetic energy.
  • Therefore, electrons are emitted with a range of kinetic energies up to the maximum defined by the equation.
  • Figure 12 Einstein’s photoelectric equation
  • Einstein’s equation indicates that there will be no photoelectric emission unless [math]hf > \phi [/math]. The frequency that is just large enough to liberate electrons, [math] f_o [/math] , is called the threshold frequency, so
  • [math] \phi = hf_0 [/math]
  • Example:
    (1)
    The work function for zinc is 4.3eV. Explain why photoelectric emission is observed when ultraviolet light of wavelength in the order of 200nm is shone onto a zinc plate but not when a 60W filament lamp is used.
    Solution:
    A photon of the ultraviolet light has energy:Given data:
    Work function [math] = \phi = 4.3 \ \text{eV} = 4.3 \times 1.6 \times 10^{-19} \ \text{J eV}^{-1} = 6.88 \times 10^{-19} \ \text{J} [/math]
    Wavelength of Ultraviolet light [math] = \lambda = 200 \times 10^{-9} \ \text{m} [/math]
    Energy for Ultraviolet light = E =? 

    Formula:

  • [math] E = hf  \\ E = h \frac{c}{\lambda} [/math]
  • Solution:
  • [math] E = h \frac{c}{\lambda} \\
    E = \frac{(6.63 \times 10^{-34}) \times (3 \times 10^8)}{200 \times 10^{-9}} \\
    E = 9.95 \times 10^{-19} \ \text{J} \\
    E = \frac{9.95 \times 10^{-19} \ \text{J}}{1.6 \times 10^{-19} \ \text{J eV}^{-1}} = 6.2 \ \text{eV} [/math]
  • This is greater than the 4.3eV work function for zinc. This means that each photon will have sufficient energy to remove an electron from the surface of the zinc and photoemission will occur.
  • The shortest wavelength (highest frequency) visible light is 400nm (at the blue end of the visible spectrum). The threshold frequency, [math]f_o[/math]  , for zinc is given by:
  • [math] \phi = hf_0 \\
    f_0 = \frac{\phi}{h} \\
    f_0 = \frac{6.88 \times 10^{-19} \ \text{J}}{6.63 \times 10^{-34} \ \text{Js}} \\
    f_0 = 1 \times 10^{15} \ \text{Hz}[/math]
  • The frequency of the blue end of the visible spectrum is:
  • [math]  f = \frac{c}{\lambda} [/math]
    [math]f = \frac{3 \times 10^8}{400 \times 10^{-9}} =  7.5 \times 10^{14} \ \text{Hz} \quad \text{(that is, } < 1 \times 10^{15} \ \text{Hz)} [/math]
  • Photoemission therefore will not take place.

9. Electron-volt:

  • An electron-volt is the work done on (or the energy gained by) an electron when it moves through a potential difference of 1 volt.
  • The energy of a photon of green light was [math] 3.7 × 10^{-19} \text{J} [/math].
  • This is a very small amount of energy, so rather than keep having to include powers of [mathh]10^{-19}[/math] when considering photon energy, we often use the electron-volt (eV) as a convenient unit of energy.
  • As an electron-volt is the work done on an electron when it moves through a potential difference of 1 volt, from
  • [math]W = QV [/math]
  • we get:
  • [math] 1 \ \text{electron-volt} = 1.6 \times 10^{-19} \ \text{C} \times 1 \ \text{V} = 1.6 \times 10^{-19} \ \text{J}[/math]

  • As the electron-volt is a very small unit of energy, we often use keV ( [math]10^3[/math] eV) and MeV ([math]10^6[/math]  eV). For example, typical X-rays have energy of 120keV and alpha particles from americium-241 (commonly found in smoke detectors) have energy of 5.6MeV.

10. Phototube:

  • A phototube is the name given to a particular type of photocell that generates photoelectrons when light falls on a specially coated metal cathode.
  • The other types of photocells are photovoltaic photocells, in which an e.m.f. is generated by the presence of light across the boundary of two semiconducting materials, and photoconductive cells, or light dependent resistors (LDRs).
  • An LDR is a semiconductor whose resistance decreases (that is, it becomes a better conductor) when it is exposed to electromagnetic radiation.
  • This is because the photon energy releases more electrons to act as charge carriers: n increases in
  • [math] I = nAvq [/math]
  • The photoelectric effect provides evidence for the particle nature of electromagnetic radiation in several ways:
    1. Threshold Frequency: The existence of a threshold frequency below which no photoelectrons are emitted, regardless of intensity, suggests that light has a particle-like behavior. Particles (photons) with energy below the threshold cannot eject electrons.
    2. Energy of Photoelectrons: The energy of photoelectrons depends on the frequency, not intensity, of light. This implies that each photon transfers its energy to an electron, consistent with particle-like behavior.
    3. Quantization: The photoelectric effect demonstrates quantization, where energy is transferred in discrete packets (photons) rather than continuously.
    4. Particle-Like Behavior: The photoelectric effect shows that light exhibits particle-like behavior, such as having energy and momentum, which is inconsistent with classical wave theory.
    5. Einstein’s Explanation: Albert Einstein’s 1905 theory, which introduced light quanta (photons), provided a comprehensive explanation for the photoelectric effect, further supporting the particle nature of light.
  • The photoelectric effect is a fundamental phenomenon that demonstrates the particle nature of electromagnetic radiation, challenging classical wave theory and leading to a deeper understanding of the behavior of light and matter.

11. Atomic Spectra:

  • Certain metal salts can be used to color a flame; sodium produces a yellow flame, potassium a lilac flame, and barium a green flame.
  • A diffraction grating shows a sequence of brilliant lines when the flame is examined through it.
  • Two yellow lines are produced by sodium, two red and two violet lines by potassium, and several lines ranging from red to violet, with green being the brightest, by barium.
  • Since light is being released, these spectra are known as emission spectra. When we determine what is causing the emission, we will define this later.
  • Kirchhoff proved that the spectra were typical of the atoms and molecules that produced them in the middle of the nineteenth century.
  • Applying a significant potential difference between the ends of a low-pressure, gas-filled tube can likewise result in the production of spectral lines.
  • These devices, which go by the name of gas discharge tubes, are ubiquitous and take the shape of street lights.
  • Due to the presence of sodium, which mostly emits deep yellow wavelengths between 589.0 and 589.6 nm, they seem yellow.
  • Because of its monochromatic output, which enhances contrast perception and minimizes light dispersion through fog and rain, this type of light is said to be safer to drive under than white light.
  • Figure 13 monochromatic light out comes from white light
  • Four lines may be distinguished when a hydrogen-filled tube is examined closely via a diffraction grating using a device known as a spectrometer: two blue lines, a bluish-green line, and a red line.
  • After analyzing the frequency of these lines, Balmer found in 1885 that a straightforward mathematical formula could be used to determine the frequency, or f, of the hydrogen lines. Rydberg changed Balmer’s equation and put it in this format:
  • [math] f = R \left( \frac{1}{n^2} – \frac{1}{m^2} \right) \\
    f = R \left( \frac{1}{2^2} – \frac{1}{m^2} \right) [/math]
  • where R is a constant (now called the Rydberg constant) and m had the values 3, 4, 5 and 6 for each of the lines in turn.
    Figure 14a shows the visible spectrum for white light. Figure 14b shows the emission spectrum of hydrogen and Figure 14c shows the absorption spectrum for hydrogen.

  • Figure 14 Hydrogen spectra

  • By applying quantum theory to his model, Bohr concluded that the electron could only exist in certain discrete or quantized orbits. This means orbits having distinct, fixed amounts of energy.
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