Particles and Waves

## Quantum Physics

Light and matter exhibit both wave-like and particle-like properties.

Photoelectric Effect: When light of sufficient frequency shines on a metal surface, electrons Are emitted.

• Electrons are emitted instantaneously, not after a delay
• No electrons are emitted if the frequency is below the threshold frequency $f_0$ Regardless of intensity
• The maximum kinetic energy of emitted electrons depends on frequency, not intensity
• More intense light produces more electrons, not more energetic ones

Classical wave theory predicts that the energy of a light wave depends on its intensity (amplitude), Not its frequency. A sufficiently intense low-frequency light should eventually eject electrons. This does not happen. Einstein’s explanation — that light consists of discrete photons with energy $E = hf$ — correctly predicts that the kinetic energy of emitted electrons depends on frequency, And that there is a threshold frequency below which no electrons are emitted regardless of intensity. This was one of the key experiments that led to quantum mechanics.

Einstein’s Photoelectric Equation:

$$E_k = hf - \phi$$

Where $\phi = hf_0$ is the work function of the metal.

Example: The work function of sodium is $2.28 \mathrm{ eV}$. Find the threshold frequency and the Maximum kinetic energy of photoelectrons when illuminated by light of frequency $8 \times 10^{14} \mathrm{ Hz}$.

Threshold frequency: $f_0 = \dfrac{\phi}{h} = \dfrac{2.28 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}} = \dfrac{3.648 \times 10^{-19}}{6.63 \times 10^{-34}} = 5.50 \times 10^{14} \mathrm{ Hz}$.

Maximum kinetic energy: $E_k = hf - \phi = 6.63 \times 10^{-34} \times 8 \times 10^{14} - 3.648 \times 10^{-19}$

$$= 5.304 \times 10^{-19} - 3.648 \times 10^{-19} = 1.656 \times 10^{-19} \mathrm{ J} = 1.04 \mathrm{ eV}$$

All matter has wave-like properties. The de Broglie wavelength of a particle with momentum $p$ is:

$$\lambda = \frac{h}{p} = \frac{h}{mv}$$

Example: Find the de Broglie wavelength of an electron accelerated through a potential Difference of $200 \mathrm{ V}$.

$$E_k = eV = 200 \mathrm{ eV} = 200 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-17} \mathrm{ J}$$ $$E_k = \frac{1}{2}mv^2 = \frac{p^2}{2m}$$ $$p = \sqrt{2mE_k} = \sqrt{2 \times 9.11 \times 10^{-31} \times 3.2 \times 10^{-17}} = \sqrt{5.83 \times 10^{-47}} = 7.64 \times 10^{-24} \mathrm{ kg m/s}$$ $$\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{7.64 \times 10^{-24}} = 8.68 \times 10^{-11} \mathrm{ m \approx 0.087 \mathrm{ nm}}$$

A $1 \mathrm{ kg}$ ball moving at $1 \mathrm{ m/s}$ has a de Broglie wavelength of $\lambda = 6.63 \times 10^{-34} / 1 = 6.63 \times 10^{-34}$ m. This is unfathomably small — far Smaller than any aperture or obstacle. Wave effects (diffraction, interference) are only observable when the wavelength is comparable to the size of the obstacles. For electrons (small mass), the de Broglie wavelength can be comparable to atomic spacing, which is why electron diffraction is readily Observable.

Electrons in atoms exist in discrete energy levels. Transitions between levels produce photons:

$$\Delta E = hf = \frac{hc}{\lambda}$$

• Emission spectrum: Bright lines on a dark background (photons emitted when electrons move to lower levels)
• Absorption spectrum: Dark lines on a continuous spectrum (photons absorbed when electrons move to higher levels)

Example: An electron in a hydrogen atom transitions from $n = 3$ to $n = 1$. The energy levels Are E_1 = -13.6 \mathrm{ eV}$$E_2 = -3.4 \mathrm{ eV}$$E_3 = -1.51 \mathrm{ eV}. Find the wavelength of the emitted photon.

$$\Delta E = E_3 - E_1 = -1.51 - (-13.6) = 12.09 \mathrm{ eV} = 1.934 \times 10^{-18} \mathrm{ J}$$ $$\lambda = \frac{hc}{\Delta E} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{1.934 \times 10^{-18}} = \frac{1.989 \times 10^{-25}}{1.934 \times 10^{-18}} = 1.028 \times 10^{-7} \mathrm{ m \approx 103 \mathrm{ nm}}$$

This is in the ultraviolet region (Lyman series).

It is fundamentally impossible to simultaneously know both the exact position and exact momentum of A particle:

$$\Delta x \cdot \Delta p \geq \frac{\hbar}{2}$$

Where $\hbar = \dfrac{h}{2\pi}$.

The uncertainty principle is not a statement about the limitations of our instruments. It is a fundamental property of nature: a particle does not have simultaneously well-defined position And momentum. This has profound consequences: it explains why electrons cannot spiral into the Nucleus (confinement to a small volume requires large momentum, preventing collapse), and it sets a Limit on the precision of any physical theory.

Example: An electron is confined to a region of width $1 \mathrm{ nm}$. What is the minimum Uncertainty in its momentum?

$$\Delta p \geq \frac{\hbar}{2\Delta x} = \frac{1.055 \times 10^{-34}}{2 \times 10^{-9}} = 5.275 \times 10^{-26} \mathrm{ kg m/s}$$

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The Standard Model classifies fundamental particles into:

Quarks (six flavours, each with an antiquark):

Leptons (six flavours, each with an antilepton):

Gauge Bosons (force carriers):

Higgs Boson: Gives particles mass via the Higgs mechanism.

The strong force between quarks increases with distance (unlike gravity and electromagnetism, which decrease with distance). This phenomenon, called colour confinement, means that separating Quarks requires more and more energy, until it becomes energetically favourable to create new Quark-antiquark pairs. The result is that quarks are always found in colour-neutral combinations: Baryons (three quarks) and mesons (quark-antiquark pair).

In all particle interactions, the following are conserved:

• Charge
• Baryon number
• Lepton number
• Energy and momentum
• Strangeness (in strong interactions)

Feynman diagrams represent particle interactions visually. Key features:

• Straight lines represent matter particles (left to right) or antimatter (right to left)
• Wavy lines represent photons
• Curly lines represent gluons
• Dashed lines represent $W$ or $Z$ bosons

Beta decay: A neutron converts to a proton by emitting a $W^-$ boson, which decays into an Electron and electron antineutrino:

$n \to p + W^-$ $W^- \to e^- + \bar{\nu}_e$

Every particle has a corresponding antiparticle with the same mass but opposite charge.

Electron-positron annihilation:

$$e^- + e^+ \to 2\gamma$$

The total energy of the photons equals $2m_e c^2$ plus any kinetic energy.

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Principle of Superposition: When two or more waves overlap, the resultant displacement at any Point is the sum of the individual displacements.

Coherent sources have the same frequency and a constant phase relationship.

A stationary (standing) wave is formed by the superposition of two progressive waves of the same frequency travelling in opposite directions.

Nodes: Points of zero amplitude.

Antinodes: Points of maximum amplitude.

String fixed at both ends:

$$f_n = \frac{nv}{2L}, \quad n = 1, 2, 3, \ldots$$

Where $L$ is the string length and $v$ is the wave speed.

Pipe open at both ends:

$$f_n = \frac{nv}{2L}, \quad n = 1, 2, 3, \ldots$$

Pipe closed at one end:

$$f_n = \frac{nv}{4L}, \quad n = 1, 3, 5, \ldots$$

Example: A guitar string of length $65 \mathrm{ cm}$ has a fundamental frequency of $330 \mathrm{ Hz}$. Find the wave speed and the frequency of the third harmonic.

$$v = 2Lf_1 = 2 \times 0.65 \times 330 = 429 \mathrm{ m/s}$$ $$f_3 = \frac{3v}{2L} = 3 \times 330 = 990 \mathrm{ Hz}$$

At the closed end, there must be a displacement node (the air cannot move). At the open end, there is a displacement antinode. The fundamental has a quarter wavelength fitting in the pipe. The second harmonic would require three-quarters of a wavelength, which gives the frequency $3f_1$. The pattern Continues with only odd multiples of the fundamental.

When a source of waves moves relative to an observer, the observed frequency changes:

$$f' = f\frac{v}{v \pm v_s}$$

Where $v_s$ is the speed of the source (minus for approaching, plus for receding).

For electromagnetic waves (relativistic):

$$f' = f\sqrt{\frac{c \pm v}{c \mp v}}$$

Example: An ambulance siren emits sound at $800 \mathrm{ Hz}$. If the ambulance approaches at $25 \mathrm{ m/s}$ (speed of sound $= 343 \mathrm{ m/s}$), find the observed frequency.

$$f' = 800 \times \frac{343}{343 - 25} = 800 \times \frac{343}{318} = 800 \times 1.0786 = 862.9 \mathrm{ Hz}$$

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Light is both a wave and a particle: Imagine light as a chameleon, it behaves like a wave (diffracting, interfering) in some experiments and like a particle (ejecting electrons) in others. This wave-particle duality is one of the most counterintuitive concepts in physics, but it’s essential for understanding the quantum world.

Why it matters: Quantum physics explains how atoms work, how lasers produce light, how semiconductors conduct electricity, and how MRI machines image our bodies. Understanding the photoelectric effect, de Broglie wavelength, and energy quantization is fundamental to modern technology.

The key insight: Energy comes in discrete packets (quanta), not continuous streams, this is why the photoelectric effect has a threshold frequency below which no electrons are emitted, regardless of light intensity.

1. Units in the photoelectric effect: Convert between eV and joules as needed. $1 \mathrm{ eV} = 1.6 \times 10^{-19} \mathrm{ J}$.

2. De Broglie wavelength of massive objects: While all matter has a de Broglie wavelength, it is negligibly small for macroscopic objects.

3. Conservation laws: Always check charge, baryon number, and lepton number are conserved in particle reactions.

4. Stationary wave harmonics: A pipe closed at one end only supports odd harmonics ($n = 1, 3, 5, \ldots$).

5. Doppler effect sign convention: Approaching sources increase observed frequency; receding sources decrease it.

6. Confusing baryon number and atomic mass number. Baryon number counts the number of quarks minus antiquarks (each quark has $B = 1/3$ Each antiquark has $B = -1/3$). A proton has $B = 1$ a neutron has $B = 1$ A meson has $B = 0$.

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1. The work function of potassium is $2.30 \mathrm{ eV}$. Find the threshold wavelength and the maximum kinetic energy of photoelectrons when illuminated by $400 \mathrm{ nm}$ light.

2. Calculate the de Broglie wavelength of a proton moving at $2 \times 10^6 \mathrm{ m/s}$.

3. Verify that the following reaction conserves charge, baryon number, and lepton number: $\pi^- + p \to K^0 + \Lambda^0$.

4. A stationary wave on a string of length $0.8 \mathrm{ m}$ has a third harmonic frequency of $600 \mathrm{ Hz}$. Find the wave speed.

5. Draw a Feynman diagram for electron-proton scattering via photon exchange.

6. A source emitting $500 \mathrm{ Hz}$ sound moves away from a stationary observer at $30 \mathrm{ m/s}$. Speed of sound is $343 \mathrm{ m/s}$. Find the observed frequency.

7. In a hydrogen atom, an electron transitions from $n = 4$ to $n = 2$. Calculate the wavelength of the emitted photon.

8. Explain how the Heisenberg uncertainty principle limits the precision of simultaneous measurements of position and momentum.

9. A neutron decays into a proton, electron, and electron antineutrino. Write the full reaction and verify conservation of charge, baryon number, and lepton number.

10. A stationary wave is formed on a string of length $1.2 \mathrm{ m}$ with a fundamental frequency of $200 \mathrm{ Hz}$. Calculate the frequencies of the second, third, and fourth harmonics, and the positions of the nodes and antinodes for the second harmonic.

When light of wavelength $450 \mathrm{ nm}$ is incident on a sodium surface, the stopping potential is Measured to be $0.65 \mathrm{ V}$. Find the work function of sodium and the threshold frequency.

$$E_k = eV_s = 1.6 \times 10^{-19} \times 0.65 = 1.04 \times 10^{-19} \mathrm{ J} = 0.65 \mathrm{ eV}$$

Photon energy: $E = hf = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{450 \times 10^{-9}} = 4.42 \times 10^{-19} \mathrm{ J} = 2.76 \mathrm{ eV}$

$$\phi = E - E_k = 2.76 - 0.65 = 2.11 \mathrm{ eV}$$

Threshold frequency: $f_0 = \frac{\phi}{h} = \frac{2.11 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}} = 5.09 \times 10^{14} \mathrm{ Hz}$

Threshold wavelength: $\lambda_0 = \frac{c}{f_0} = \frac{3 \times 10^8}{5.09 \times 10^{14}} = 5.89 \times 10^{-7} \mathrm{ m} = 589 \mathrm{ nm}$

This is in the yellow part of the visible spectrum, so sodium only emits photoelectrons when illuminated with blue, violet, or UV light.

A metal surface with work function $2.0 \mathrm{ eV}$ is illuminated with light of frequency $7 \times 10^{14} \mathrm{ Hz}$ at an intensity of $5 \mathrm{ W/m^2}$. The surface area is $2 \mathrm{ cm^2}$.

Photon energy: $E = hf = 6.63 \times 10^{-34} \times 7 \times 10^{14} = 4.64 \times 10^{-19} \mathrm{ J} = 2.90 \mathrm{ eV}$

Since $2.90 \mathrm{ eV \gt 2.0 \mathrm{ eV}}$ Photoelectrons are emitted.

Maximum KE: $E_k = 2.90 - 2.0 = 0.90 \mathrm{ eV}$

Photon flux: Power per unit area divided by energy per photon:

$$\mathrm{flux} = \frac{5}{4.64 \times 10^{-19}} = 1.078 \times 10^{19} \mathrm{ photons/m^2\mathrm{/s}}$$

Photoelectrons per second: $1.078 \times 10^{19} \times 2 \times 10^{-4} = 2.16 \times 10^{15} \mathrm{ electrons/s}$

Maximum current: $I = ne = 2.16 \times 10^{15} \times 1.6 \times 10^{-19} = 3.45 \times 10^{-4} \mathrm{ A} = 0.345 \mathrm{ mA}$

Electrons are accelerated through a potential difference of $500 \mathrm{ V}$. They pass through a thin crystal and produce a diffraction pattern. The first diffraction maximum is observed at an angle of $1.8^{\circ}$. Calculate the atomic spacing.

$$E_k = eV = 500 \mathrm{ eV} = 8.0 \times 10^{-17} \mathrm{ J}$$ $$p = \sqrt{2mE_k} = \sqrt{2 \times 9.11 \times 10^{-31} \times 8.0 \times 10^{-17}} = \sqrt{1.458 \times 10^{-46}} = 1.208 \times 10^{-23} \mathrm{ kg m/s}$$ $$\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{1.208 \times 10^{-23}} = 5.49 \times 10^{-11} \mathrm{ m}$$

For the first-order maximum: $d\sin\theta = \lambda$

$$d = \frac{\lambda}{\sin\theta} = \frac{5.49 \times 10^{-11}}{\sin 1.8^{\circ}} = \frac{5.49 \times 10^{-11}}{0.0314} = 1.75 \times 10^{-9} \mathrm{ m} = 1.75 \mathrm{ nm}$$

This is roughly 3—5 atomic spacings, which is consistent with crystal lattice spacing.

Calculate the wavelength of the first three lines in the Balmer series (transitions to $n = 2$).

$E_n = -\frac{13.6}{n^2} \mathrm{ eV}$

$n = 3 \to 2$:

$$\Delta E = 13.6\left(\frac{1}{4} - \frac{1}{9}\right) = 13.6 \times \frac{5}{36} = 1.889 \mathrm{ eV}$$ $$\lambda = \frac{hc}{\Delta E} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{1.889 \times 1.6 \times 10^{-19}} = \frac{1.989 \times 10^{-25}}{3.022 \times 10^{-19}} = 6.58 \times 10^{-7} \mathrm{ m} = 658 \mathrm{ nm \mathrm{ (red)}}$$

$n = 4 \to 2$:

$$\Delta E = 13.6\left(\frac{1}{4} - \frac{1}{16}\right) = 13.6 \times \frac{3}{16} = 2.55 \mathrm{ eV}$$ $$\lambda = \frac{1.989 \times 10^{-25}}{2.55 \times 1.6 \times 10^{-19}} = 4.87 \times 10^{-7} \mathrm{ m} = 487 \mathrm{ nm \mathrm{ (blue}}-green)$$

$n = 5 \to 2$:

$$\Delta E = 13.6\left(\frac{1}{4} - \frac{1}{25}\right) = 13.6 \times \frac{21}{100} = 2.856 \mathrm{ eV}$$ $$\lambda = \frac{1.989 \times 10^{-25}}{2.856 \times 1.6 \times 10^{-19}} = 4.35 \times 10^{-7} \mathrm{ m} = 435 \mathrm{ nm \mathrm{ (violet)}}$$

Verify conservation of charge, baryon number, and lepton number for beta-minus decay of a free Neutron:

$$n \to p + e^- + \bar{\nu}_e$$

Writing with full quark content: $udd \to uud + e^- + \bar{\nu}_e$

A $\pi^-$ meson (quark content $d\bar{u}$) decays: $\pi^- \to \mu^- + \bar{\nu}_\mu$

A string of length $1.2 \mathrm{ m}$ has fundamental frequency $200 \mathrm{ Hz}$.

Wave speed: $v = 2Lf_1 = 2 \times 1.2 \times 200 = 480 \mathrm{ m/s}$

Second harmonic ($n = 2$): $f_2 = 400 \mathrm{ Hz}$, $\lambda_2 = \frac{2L}{2} = 1.2 \mathrm{ m}$.

Nodes at: $0, 0.6 \mathrm{ m, 1.2 \mathrm{ m}}$ (3 nodes, including the fixed ends)

Antinodes at: $0.3 \mathrm{ m, 0.9 \mathrm{ m}}$ (2 antinodes)

Third harmonic ($n = 3$): $f_3 = 600 \mathrm{ Hz}$, $\lambda_3 = \frac{2L}{3} = 0.8 \mathrm{ m}$.

Fourth harmonic ($n = 4$): $f_4 = 800 \mathrm{ Hz}$, $\lambda_4 = \frac{2L}{4} = 0.6 \mathrm{ m}$.

1. The work function of caesium is $2.14 \mathrm{ eV}$. Calculate the threshold wavelength and the maximum KE of photoelectrons when illuminated with $550 \mathrm{ nm}$ light.

2. Calculate the de Broglie wavelength of a neutron moving at $2200 \mathrm{ m/s}$ (thermal neutrons in a nuclear reactor). (Mass of neutron $= 1.675 \times 10^{-27} \mathrm{ kg}$.)

3. A hydrogen atom is in the $n = 4$ state. Calculate the wavelengths of all possible photons emitted as it decays to the ground state.

4. Verify conservation of charge, baryon number, and lepton number for the reaction: $\pi^+ + p \to K^+ + \Sigma^+$

5. A stationary wave is set up on a string of length $0.6 \mathrm{ m}$ with a fundamental frequency of $250 \mathrm{ Hz}$. Calculate the wave speed and the frequency of the fifth harmonic.

6. A source emitting $600 \mathrm{ Hz}$ moves towards a stationary observer at $40 \mathrm{ m/s}$. Speed of sound $= 343 \mathrm{ m/s}$. Calculate the observed frequency and the wavelength of the observed sound.

7. Explain how electron diffraction experiments provide evidence for the wave nature of matter.

8. Calculate the energy of a photon in the Lyman series corresponding to a transition from $n = 5$ to $n = 1$ in hydrogen.

9. An electron is confined to a region of width $0.5 \mathrm{ nm}$. Estimate the minimum uncertainty in its velocity.

10. Explain why the strong nuclear force must be a short-range force. Reference colour confinement and the fact that quarks are never observed in isolation.

An electron is accelerated through a potential difference of $200 \mathrm{ V}$. Calculate its de Broglie wavelength.

Step 1: Find the kinetic energy

$$eV = \frac{1}{2}mv^2$$ $$1.602 \times 10^{-19} \times 200 = \frac{1}{2} \times 9.109 \times 10^{-31} \times v^2$$ $$v^2 = \frac{2 \times 3.204 \times 10^{-17}}{9.109 \times 10^{-31}} = 7.034 \times 10^{13}$$ $$v = 8.387 \times 10^6 \mathrm{ m/s}$$

Step 2: Calculate the de Broglie wavelength

$$\lambda = \frac{h}{mv} = \frac{6.626 \times 10^{-34}}{9.109 \times 10^{-31} \times 8.387 \times 10^6}$$ $$\lambda = \frac{6.626 \times 10^{-34}}{7.639 \times 10^{-24}} = 8.67 \times 10^{-11} \mathrm{ m} = 0.0867 \mathrm{ nm}$$