QUANTUM PHYSICS

One of the most interesting topics in Physics and especially A level physics is  A level physics Quantum Physics.Lets embark your learning with specification targeted content below:

1. Photon

1. a) The particulate nature (photon model) of electromagnetic radiation:

2. The particulate nature of electromagnetic radiation, also known as the photon model, is a fundamental concept in physics that describes the behavior of electromagnetic radiation as particles called photons.
3. In quantum mechanics, a photon is a particle that represents a quantum of energy of electromagnetic radiation. This means that photons are the smallest units of energy that can be exchanged between electromagnetic fields and matter.
4. Here are some key aspects of the photon model:
5. – Photon as a particle: Photons are massless particles that have energy and momentum, but no electric charge.
6. – Quantization of energy: Photons have discrete energy packets (quanta) that depend on their frequency (where E is energy, h is Planck’s constant, and f is frequency).
7. – Wave-particle duality: Photons exhibit both wave-like (diffraction, interference) and particle-like (particle-like behavior in interactions) behavior.
8. – Particle-like behavior: Photons interact with matter as particles, exhibiting phenomena like the photoelectric effect, Compton scattering, and pair production.
9. – Zero mass: Photons have no rest mass, but they do have momentum (where p is momentum, E is energy, and c is the speed of light).
10. – Speed of light: Photons always travel at the speed of light (c) in a vacuum.
11. The photon model is a cornerstone of quantum mechanics and has been extensively experimentally confirmed. It’s a fundamental concept in understanding the behavior of electromagnetic radiation and its interactions with matter.

c) Energy of a photon:

• The energy of a single photon is given by the equation
• [math]E = hf[/math]
• Where E is the photon energy in J, h is the Planck constant and f is the frequency of the radiation in Hz.
• The Planck constant,h, has a value of [math]6.626 \times 10^{-34} \, \text{Js}[/math].
• The energy of a photon is directly proportional to its frequency, and photon energies are always emitted in whole number multiples:[math]E = n hf \quad (n = 0, 1, 2, 3, \dots)[/math]
• The equation relating the speed of light (c) the frequency of a wave (f) and its wavelength (λ) is wave
• [math]\text{speed} = \text{frequency} \times \text{wavelength}[/math]
• or
• [math]c = f \lambda[/math]
• [math]f = \frac{c}{\lambda}[/math]
• Rearranging this equation and substituting into
• [math]E = hf[/math]
• Gives an alternative equation for calculating the energy of a photon:
• [math]E = \frac{h c}{\lambda}[/math]
• If we know the wavelength of the radiation.
• Example:
• Calculate the energy of a photon with:
• (a) a frequency of[math]4.6 \times 10^{18} \, \text{Hz}[/math]
• (b) a wavelength of [math]4.2 \times 10^{-8}[/math]m. The speed of light is[math]3 \times 10^{8} \, \text{m/s}[/math].
• Given data:
• Frequency = f =[math]4.6 \times 10^{18} \, \text{Hz}[/math]
• Wavelength =[math]\lambda[/math] =  [math]4.2 \times 10^{-8}[/math]m
• Speed of light = c =[math]3 \times 10^8 \, \text{m/s}[/math]
• Find data:
• Energy of photon = E =?
• Energy of photon for m wavelength = E =?
• Formula:
• a)[math]E = hf[/math]
• b)[math]E = \frac{h c}{\lambda}[/math]
• Solution:
• a).
• [math]E = hf[/math]
• [math]E = (6.626 \times 10^{-34})(4.6 \times 10^{18})[/math]
• [math]E = 3.0 \times 10^{-15} \, \text{J}[/math]
• b).
• [math]E = \frac{h c}{\lambda}[/math]
• [math]E = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{4.6 \times 10^{18}}[/math]
• [math]E = 4.7 \times 10^{-18} \, \text{J}[/math]

d) The electron volt (eV) as a unit of energy:

• The electron volt (eV) is a unit of energy that represents the energy gained by a single electron when it is accelerated through a potential difference of one volt. It is a commonly used unit of energy in physics, particularly in the fields of atomic and subatomic physics.
• [math]1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{Joules (J)}[/math]
• Properties:
• – It is a small unit of energy, suitable for describing the energy of individual electrons or photons.
• – It is a convenient unit for expressing the energy of atomic and subatomic particles.
• – It is used to describe the energy of electromagnetic radiation, such as photons.
• Examples:
• The energy of a photon of visible light is typically around 1-10 eV.
• The energy of an electron in a hydrogen atom is typically around 10-100 eV.
• The energy of a high-energy particle in a particle accelerator can be millions or even billions of eV.
• Conversions:
• [math]1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}[/math]
• [math]1 \, \text{J} = 6.242 \times 10^{18} \, \text{eV}[/math]
• The electron volt is a non-SI unit, but it is widely used in physics and is accepted as a convenient unit of energy.
• 
• Figure 1 The kinetic energy of an electron will increase by 1 eV when it is accelerated across a potential difference of 1 V. (a) Shows an electron leaving the negative plate. (b) Shows the electron arriving at the positive plate having gained 1 eV of kinetic energy.
• We calculate the energy gained by an electron (in Figure 1) using:
• [math]\text{Energy gained} = \text{Charge} \times \text{Potential difference}[/math]
• or
• [math]E=QV[/math]
• The charge on an electron is [math]1.602 \times 10^{-19}[/math]C and the potential difference is 1 V,
• so:
• [math]E = 1.602 \times 10^{-19} \, \text{C} \times 1 \, \text{V} = 1.602 \times 10^{-19} \, \text{J}[/math]
• This means that 1 eV is equal to[math]1.602 \times 10^{-19}[/math]J
• If we increase the particle’s charge or the potential difference between the plates, then the kinetic energy acquired by the particle will increase in direct proportion.
• So, if an electron of charge e is accelerated through a potential difference of 100 V it will gain 100 eV of kinetic energy.

e) I) Using LEDs and the equation [math][/math] to estimate the value of Planck constant h.

• Using LEDs and the equation eV = hc/λ, you can estimate the value of Planck’s constant (h). Here’s a brief outline of the experiment:
• Materials:
• – LEDs of different colors (e.g., red, orange, yellow, green, blue, violet)
• – A multimeter (for measuring voltage)
• – A spectrometer (for measuring wavelength)
• Procedure:
• – Measure the voltage across each LED using a multimeter.
• – Measure the wavelength of light emitted by each LED using a spectrometer.
• – Use the equation eV = hc/λ to calculate the energy (eV) of each LED.
• – Plot a graph of energy (eV) vs. wavelength (λ) for each LED.
• – Use the slope of the graph to estimate the value of Planck’s constant (h).
• The equation eV = hc/λ can be rearranged to solve for h:
• [math]h = \frac{eV \times \lambda}{c}[/math]
• Where h is Planck’s constant, eV is the energy, λ is the wavelength, and c is the speed of light.
• By measuring the voltage and wavelength of each LED, you can calculate the energy (eV) and then use the equation above to estimate the value of h.
• This experiment provides an estimate of Planck’s constant, but it may not be as precise as other methods. However, it’s a great way to illustrate the relationship between energy, wavelength, and Planck’s constant.

• II) Determine the Planck constant using different colored LEDs

• To determine the Planck constant (h) using different colored LEDs, you can follow these steps:
• – Measure the wavelength (λ) of each LED using a spectrometer.
• – Measure the voltage (V) across each LED using a multimeter.
• – Calculate the energy (eV) of each LED using the formula: eV = V
• Use the formula:
• [math]h = \frac{eV \times \lambda}{c}[/math]
• To calculate Planck’s constant (h) for each LED.
• c is the speed of light (approximately [math]3 \times 10^8[/math]m/s)
• eV is the energy in electron-volts (1 eV =[math]1.602 \times 10^{-19}[/math] J)
• λ is the wavelength in meters
• Using different colored LEDs will provide a range of wavelengths and energies, allowing you to calculate Planck’s constant (h) for each LED. By plotting the values and calculating the average, you can determine an estimated value of h.
• This method provides an estimate of Planck’s constant, and the accuracy may vary depending on the equipment and measurements.
• However, it’s a fun and educational experiment to illustrate the relationship between energy, wavelength, and Planck’s constant.

The photoelectric effect:

a) Photoelectric effect, including a simple experiment to demonstrate this effect:

• The photoelectric effect is a phenomenon where light (photons) hitting a material can eject electrons from the material. This effect is a result of the particle-like behavior of light and the wave-like behavior of electrons.
• ⇒Demonstrating the photoelectric effect with a gold-leaf electroscope:
• The gold-leaf electroscope is composed of a brass stem to which a thin gold leaf is attached.
• There is a metal cap attached to the top of the stem and the metal to be irradiated with electromagnetic radiation is laid on the metal cap.
• 
• Figure 2 The photoelectric effect can be demonstrated using a gold-leaf electroscope.
• A metal plate (usually zinc) is placed on the metal and is then charged negatively by touching it with a negatively charged polythene rod, or by electrostatic induction.
• When this is done, the metal stem and gold leaf will also become negatively charged, meaning that the stem and the leaf will repel each other.
• It is also possible to make the zinc plate, metal stem and the gold leaf positively charged.
• ⇒Observations of the photoelectric effect:
• The photoelectric effect is investigated for two different sources of radiation – visible light from a standard desktop lamp and UV light from a UV light source (Figure 3).
• When visible light is incident on a positively charged metal plate, there is no movement of the gold leaf. The same is also true when UV light is shone on the positively charged plate.
• 
• Figure 3 The photoelectric effect is investigated for two different sources of radiation – visible light from a standard desktop lamp and UV light from a UV light source
• Shining visible light on the negatively charged zinc plate also causes no movement of the gold leaf. No matter how bright or intense the visible light beam is, no movement of the gold leaf is seen.
• However, when UV light is shone on the negatively charged zinc plate, the gold leaf falls. This shows that the metal plate loses its negative charge through the emission of electrons, which are repelled by the negative charge on the electroscope.
• The discharge of the electroscope cannot be caused by ions in the air, because the electroscope is in a sealed vacuum.

c) Einstein’s photoelectric equation[math]hf = \phi + \text{K.E.}_{\text{max}}[/math] :

• Einstein also suggested that each single photon could only eject one electron from the metal surface – either the photon energy was larger than the work function of the metal and the electron would be released with some kinetic energy.
• The photon would not have enough energy and the emitted electron would stay on the metal surface. Applying the principle of conservation of energy, Einstein suggested an equation to explain the photoelectric effect:
• [math]hf = \phi + \text{K.E.}_{\text{max}}[/math]
• Where:
• [math]hf[/math]= The photon energy of an incident photon
• [math]h[/math]= Plank’s constant
• [math] \phi [/math]= The work function of the metal
• [math] \text{K.E.}_{\text{max}}[/math]= The maximum kinetic energy of an electron once it has been ejected from the surface of the metal.
• This equation explains why there is a threshold frequency for emission of photoelectrons, and shows that light can behave like a stream of particles.
• The photoelectric effect cannot be explained using a wave model since in the wave model, the energy of a wave depends on its amplitude (intensity) not its frequency.
• Einstein’s detailed explanation of the photoelectric effect led to him winning the Nobel Prize for physics in 1921.
• 
• Figure 4 Changing the conditions for photoelectron emission.

d) Work function; threshold frequency

• Einstein’s explanation for the photoelectric effect was that in order for electrons to be released from a metal, the frequency of the incident radiation must exceed the threshold frequency for that metal. This is needed to provide at least the minimum energy required to release an electron from the surface, this minimum energy the work function of the metal.
• Figure 4 can be interpreted as,
• The incident radiation is below the threshold frequency – this means the photons do not have enough energy to overcome the work function of the metal.
• The incident radiation has a frequency equal to the threshold frequency – this will cause electrons to be omitted from the metal surface, but with zero kinetic energy.
• The incident radiation is above the threshold frequency – this will cause electrons to be ejected with some kinetic energy Increasing the frequency further increases the kinetic energy of the photoelectrons.
• Increasing the incident wave intensity (of a single frequency that is above the threshold frequency) ejects more electrons from the metal surface per second, but it will not affect their kinetic energy.
• Electrons will be emitted from the surface of a metal only if the incident radiation is above a minimum frequency called the threshold frequency. This is determined by the metal itself and is different for different metals. Below the threshold frequency, no electrons are emitted.
• Emission of electrons starts the instant the surface starts to be irradiated, provided that the incident radiation exceeds the threshold frequency.

e) The idea that the maximum kinetic energy of the photoelectrons is independent of the intensity of the incident radiation:

• Figure 6 shows a photocell circuit used to investigate the energy of the emitted electrons.
• The frequency of the incident radiation is above the threshold frequency for this metal.
• The photocell is enclosed in a vacuum so that there are no collisions between the electrons and gas molecules in the air. As shown in Figure 1, plate A is connected to the positive terminal of the power supply.
• 
• Figure 6 Measuring electron kinetic energy
• The p.d. across the photocell is initially zero.
• The photoelectrons that are emitted from A due to the incident UV radiation cross the gap and this will register as a current on the microammeter.
• If the potential divider is adjusted, then the potential difference between A and B is increased and electrons emitted from A are attracted back to A.
• The current decreases until, eventually, even the most energetic photoelectrons are prevented from leaving plate A and the microammeter reading is zero (Figure 7).
• 
• Figure 7 How the photoelectric current depends on the p.d. between the plates shown in Figure 6.
• The p.d. at which this occurs is called the stopping potential.We can equate the work done to accelerate or decelerate charged particles with the kinetic energy transferred to the electrons, giving:
• [math]eV = \frac{1}{2} mv^2[/math]
• At the stopping potential, all the emitted electrons have been brought to rest, so we obtain a value for the maximum kinetic energy of the photoelectrons by multiplying the stopping potential (V₀) by the charge on the electron (e) .
• [math]eV_0 = \text{KE}_{\text{max}}[/math]
• So, if the stopping potential is found to be 1.83 V, the maximum kinetic energy of the photoelectrons must be 1.83V or [math]2.93 \times 10^{-19} \, \text{J}[/math]
• 
• Figure 8 Stopping potential different intensity of UV radiation
• Figure 8 shows that the stopping potential does not depend on the intensity of the incident light.

f) The idea that rate of emission of photoelectrons above the threshold frequency is directly proportional to the intensity of the incident radiation:

• The circuit in Figure 1 can also be used to investigate the rate of emission of photoelectrons.
• 
  Figure 9 to investigate the rate of emission of photoelectrons
• The current in the circuit depends on the number of photoelectrons emitted each second.
• For incident radiation of a single frequency above the threshold frequency, observations show that doubling the light intensity doubles the number of electrons emitted, but does not affect the energies of the emitted electrons.
• – The number of emitted photoelectrons is directly proportional to the intensity of the incident radiation, provided that the radiation is of a single frequency and is above the threshold frequency.
• – The kinetic energy of the emitted photoelectrons is not affected by the intensity of the incident radiation.
• – The kinetic energy of the emitted photoelectrons is affected by the frequency of the incident photons. If the frequency of the incident radiation is increased, the kinetic energy of the photoelectrons also increases.

Wave particle duality:

a) Electron diffraction, including experimental evidence of this effect.

• In 1923, the French physicist Louis de Broglie suggested something that, at the time, sounded totally preposterous.
• He claimed that all matter, regardless of its mass, could display wave-like properties.
• Evidence for this suggestion was provided by an experiment which showed that electrons showed wave-like behavior as well as behaving like particles.
• To prove that particles can also act as waves, you have to show the particles exhibiting a wave-like characteristic or property, such as diffraction or interference.
• For example, electrons can be diffracted just as light can be – a wave property.
• The apparatus used for such an experiment is shown in Figure 10.
• The wavelength of electrons is much smaller than that of light.
• The slit spacing of a diffraction grating is very large in comparison with an electron’s wavelength, so the spacing between atoms – which is very similar to the electron’s wavelength – is used instead.
• 
• Figure 10 Apparatus used to show electron diffraction
• Electrons from an electron gun are accelerated through a vacuum towards a layer of polycrystalline graphite.
• 
  Figure 11 An electron diffraction pattern.
• A polycrystalline material is made up of many tiny crystals, each consisting of a large number of regularly arranged atoms.
• The electrons diffract as they emerge from the gaps between the atomic layers in the graphite film, and interfere constructively.
• Since the graphite atoms are not all lined up in the same direction as in a diffraction grating, this gives a circular pattern instead of the parallel lines seen when light diffracts.
• An electron diffraction pattern is shown in Figure 11. The image seen on the fluorescent screen is created when electrons striking the screen cause light to be emitted.

c) The de Broglie equation:

• De Broglie proposed that if electrons and other particles travel through space as a wave, they have an associated wavelength.
• By combining the idea of an energy quantum with [math]E = mc^2[/math] he derived a formula for the wavelength, λ of a particle:
• [math]\lambda = \frac{h}{mv}[/math]
• Where m is the mass of the particle and h is the Planck constant. This is known as the de Broglie equation – mv is momentum which is often given the symbol p, so this equation is also expressed as
• [math]\lambda = \frac{h}{p}[/math]
• In 1927, two American physicists called Davisson and Germer confirmed de Broglie’s equation by observing the behavior of electrons that had been diffracted from the surface of a nickel crystal.
• By accelerating electrons, of charge e, through a potential difference of V, they observed a pattern of electron diffraction (Figure 2), from which they could calculate the electron’s wavelength.
• The predicted wavelength is found by equating the work done to accelerate the electrons with the kinetic energy transferred to the electrons.
• Substituting
• [math]eV = \frac{1}{2} mv^2[/math]
• Into the de Broglie equation gives:
• [math]\lambda = \frac{h}{\sqrt{2mVe}}[/math]
• Davisson and Germer found that their values for the wavelength of electrons from measurements of the diffraction pattern were similar to those predicted by de Broglie’s equation.
• De Broglie’s equation also predicts that the wavelength of the electron can be changed by varying its velocity.
• As Davisson and Germer increased the accelerating voltage, the rings in the diffraction pattern got bigger showing the wavelength must be getting smaller as the electrons move faster.
• 
• Figure 12 The arrangement used by Davisson and Germer
• Electron diffraction can be used to determine atomic spacing. It is also useful for determining other information about the structure of matter.
• Increasing the speed of the electrons results in a decrease in their wavelength, and so much smaller values of inter-atomic spacing can be measured.
• Today, high-speed electrons can be used to determine the arrangement of atoms in crystalline structures or to measure the diameter of a nucleus.