Particles and Waves

This chapter covers Advanced Higher Physics content, extending beyond Higher level.

Quantum Physics

Wave-Particle Duality

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

Why the Photoelectric Effect Disproves the Wave Theory of Light

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

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

De Broglie Wavelength

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 $200 \mathrm{ V$ .

$E_k = eV = 200 \mathrm{ eV = 200 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-17} \mathrm{ J$

$E_k = \frac{1}{2}mv^2 = \frac{p^2}{2m}$

$p = \sqrt{2mE_k} = \sqrt{2 \times 9.11 \times 10^{-31} \times 3.2 \times 10^{-17}} = \sqrt{5.83 \times 10^{-47}} = 7.64 \times 10^{-24} \mathrm{ kg m/s$

$\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{7.64 \times 10^{-24}} = 8.68 \times 10^{-11} \mathrm{ m \approx 0.087 \mathrm{ nm$

Why Macroscopic Objects Do Not Show Wave Behaviour

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.

Energy Levels and Spectra

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).

Heisenberg Uncertainty Principle

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

Why the Uncertainty Principle Is Not About Measurement Error

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?

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

Particle Physics

Standard Model

The Standard Model classifies fundamental particles into:

Quarks (six flavours, each with an antiquark):

Generation	Up-type	Charge	Down-type	Charge
1	Up (u)	$+2/3$	Down (d)	$-1/3$
2	Charm (c)	$+2/3$	Strange (s)	$-1/3$
3	Top (t)	$+2/3$	Bottom (b)	$-1/3$

Leptons (six flavours, each with an antilepton):

Generation	Charged	Charge	Neutrino	Charge
1	Electron (e)	$-1$	Electron neutrino	$0$
2	Muon ( $\mu$ )	$-1$	Muon neutrino	$0$
3	Tau ( $\tau$ )	$-1$	Tau neutrino	$0$

Gauge Bosons (force carriers):

Force	Boson	Mass	Acts on
Electromagnetic	Photon ( $\gamma$ )	0	Charged particles
Strong	Gluon (g)	0	Quarks, gluons
Weak	$W^+, W^-, Z^0$	Heavy	All fermions
Gravity	Graviton (hypothetical)	0	All particles

Higgs Boson: Gives particles mass via the Higgs mechanism.

Why Quarks Are Never Found in Isolation

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).

Conservation Laws in Particle Physics

In all particle interactions, the following are conserved:

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

Feynman Diagrams

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:

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

Antimatter

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.

Wave Phenomena (Advanced Higher)

Superposition and Interference

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.

Stationary Waves

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.

$v = 2Lf_1 = 2 \times 0.65 \times 330 = 429 \mathrm{ m/s$

$f_3 = \frac{3v}{2L} = 3 \times 330 = 990 \mathrm{ Hz$

Why a Closed Pipe Only Supports Odd Harmonics

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.

Doppler Effect

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.

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

Common Pitfalls

Units in the photoelectric effect: Convert between eV and joules as needed. $1 \mathrm{ eV = 1.6 \times 10^{-19} \mathrm{ J$ .
De Broglie wavelength of massive objects: While all matter has a de Broglie wavelength, it is negligibly small for macroscopic objects.
Conservation laws: Always check charge, baryon number, and lepton number are conserved in particle reactions.
Stationary wave harmonics: A pipe closed at one end only supports odd harmonics ( $n = 1, 3, 5, \ldots$ ).
Doppler effect sign convention: Approaching sources increase observed frequency; receding sources decrease it.
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 $B = 0$ .

Practice Questions

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.
Calculate the de Broglie wavelength of a proton moving at $2 \times 10^6 \mathrm{ m/s$ .
Verify that the following reaction conserves charge, baryon number, and lepton number: $\pi^- + p \to K^0 + \Lambda^0$ .
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.
Draw a Feynman diagram for electron-proton scattering via photon exchange.
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.
In a hydrogen atom, an electron transitions from $n = 4$ to $n = 2$ . Calculate the wavelength of the emitted photon.
Explain how the Heisenberg uncertainty principle limits the precision of simultaneous measurements of position and momentum.
A neutron decays into a proton, electron, and electron antineutrino. Write the full reaction and verify conservation of charge, baryon number, and lepton number.
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.

11. Photoelectric Effect: Extended Worked Examples

Worked Example: Stopping Potential

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.

Worked Example: Photoelectric Current

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

12. De Broglie Wavelength: Extended Examples

Worked Example: Electron Diffraction

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.

13. Energy Levels: Extended Analysis

Worked Example: Hydrogen Spectral Series

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

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

$n = 3 \to 2$ :

$\Delta E = 13.6\left(\frac{1}{4} - \frac{1}{9}\right) = 13.6 \times \frac{5}{36} = 1.889 \mathrm{ eV$

$\lambda = \frac{hc}{\Delta E} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{1.889 \times 1.6 \times 10^{-19}} = \frac{1.989 \times 10^{-25}}{3.022 \times 10^{-19}} = 6.58 \times 10^{-7} \mathrm{ m = 658 \mathrm{ nm \mathrm{ (red)$

$n = 4 \to 2$ :

$\Delta E = 13.6\left(\frac{1}{4} - \frac{1}{16}\right) = 13.6 \times \frac{3}{16} = 2.55 \mathrm{ eV$

$\lambda = \frac{1.989 \times 10^{-25}}{2.55 \times 1.6 \times 10^{-19}} = 4.87 \times 10^{-7} \mathrm{ m = 487 \mathrm{ nm \mathrm{ (blue-green)$

$n = 5 \to 2$ :

$\Delta E = 13.6\left(\frac{1}{4} - \frac{1}{25}\right) = 13.6 \times \frac{21}{100} = 2.856 \mathrm{ eV$

$\lambda = \frac{1.989 \times 10^{-25}}{2.856 \times 1.6 \times 10^{-19}} = 4.35 \times 10^{-7} \mathrm{ m = 435 \mathrm{ nm \mathrm{ (violet)$

14. Particle Physics: Extended Conservation Laws

Worked Example: Verifying Conservation Laws

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: $udd \to uud + e^- + \bar{\nu}_e$

Quantity	Before	After	Conserved?
Charge	$0 + (-\frac{1}{3}) + (-\frac{1}{3}) + (-\frac{1}{3}) = -1$	$\frac{2}{3} + \frac{2}{3} + (-\frac{1}{3}) + (-1) + 0 = -1$	Yes
Baryon number	$3 \times \frac{1}{3} = 1$	$3 \times \frac{1}{3} + 0 + 0 + 0 = 1$	Yes
Lepton number	$0$	$1 + (-1) = 0$	Yes

Worked Example: Pion Decay

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

Quantity	Before	After	Conserved?
Charge	$(-\frac{1}{3}) + (-\frac{2}{3}) = -1$	$-1 + 0 = -1$	Yes
Baryon number	$0 + 0 = 0$	$0 + 0 = 0$	Yes
Lepton number	$0$	$1 + (-1) = 0$	Yes

15. Stationary Waves: Extended Analysis

Worked Example: Nodes and Antinodes for the Second Harmonic

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

16. Summary Table: Quantum and Wave Formulas

Topic	Formula	Variables	Notes
Photon energy	$E = hf = hc/\lambda$	$h$ , $f$ , $\lambda$	Planck’s constant
Photoelectric	$E_k = hf - \phi$	$h$ , $f$ , $\phi$	Einstein’s equation
De Broglie	$\lambda = h/p = h/(mv)$	$h$ , $p$ , $m$ , $v$	Matter waves
Uncertainty	$\Delta x \Delta p \ge \hbar/2$	$\Delta x$ , $\Delta p$	Fundamental limit
Energy levels	$\Delta E = hf = hc/\lambda$	$h$ , $f$ , $\lambda$	Spectral lines
Standing wave	$f_n = nv/(2L)$	$n$ , $v$ , $L$	String fixed at ends
Closed pipe	$f_n = nv/(4L)$	$n$ odd	Only odd harmonics
Doppler (sound)	$f' = fv/(v \pm v_s)$	$f$ , $v$ , $v_s$	Approaching/receding

17. Practice Questions (Additional)

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.
Calculate the de Broglie wavelength of a neutron moving at $2200 \mathrm{ m/s$ (thermal neutrons in a nuclear reactor). (Mass of neutron $= 1.675 \times 10^{-27} \mathrm{ kg$ .)
A hydrogen atom is in the $n = 4$ state. Calculate the wavelengths of all possible photons emitted as it decays to the ground state.
Verify conservation of charge, baryon number, and lepton number for the reaction: $\pi^+ + p \to K^+ + \Sigma^+$
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.
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.
Explain how electron diffraction experiments provide evidence for the wave nature of matter.
Calculate the energy of a photon in the Lyman series corresponding to a transition from $n = 5$ to $n = 1$ in hydrogen.
An electron is confined to a region of width $0.5 \mathrm{ nm$ . Estimate the minimum uncertainty in its velocity.
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.

Extended Worked Examples

Example 21: De Broglie Wavelength of an Electron in a Potential Difference

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

$\lambda = \frac{h}{mv} = \frac{6.626 \times 10^{-34}}{9.109 \times 10^{-31} \times 8.387 \times 10^6}$

$\lambda = \frac{6.626 \times 10^{-34}}{7.639 \times 10^{-24}} = 8.67 \times 10^{-11} \mathrm{ m = 0.0867 \mathrm{ nm$

:::info For electrons accelerated through potential $V$ The de Broglie wavelength can be calculated Directly:

$\lambda = \frac{h}{\sqrt{2m_e eV}} = \frac{1.226}{\sqrt{V}} \mathrm{ nm$

Where $V$ is in volts. This is a very useful shortcut: for $V = 200 \mathrm{ V$ $\lambda = 1.226/\sqrt{200} = 0.0867 \mathrm{ nm$ . :::

Example 22: Photon Energy and Wavelength Relationships

A photon has energy $4.5 \mathrm{ eV$ . Calculate (a) its wavelength, (b) its frequency, and (c) its Momentum.

Step 1: Convert energy to joules

$E = 4.5 \mathrm{ eV = 4.5 \times 1.602 \times 10^{-19} = 7.209 \times 10^{-19} \mathrm{ J$

Step 2: Wavelength

$\lambda = \frac{hc}{E} = \frac{6.626 \times 10^{-34} \times 3.0 \times 10^8}{7.209 \times 10^{-19}}$

$\lambda = \frac{1.988 \times 10^{-25}}{7.209 \times 10^{-19}} = 2.758 \times 10^{-7} \mathrm{ m = 275.8 \mathrm{ nm$

This is in the ultraviolet region.

Step 3: Frequency

$f = \frac{c}{\lambda} = \frac{3.0 \times 10^8}{2.758 \times 10^{-7}} = 1.088 \times 10^{15} \mathrm{ Hz$

Step 4: Momentum

$p = \frac{E}{c} = \frac{7.209 \times 10^{-19}}{3.0 \times 10^8} = 2.403 \times 10^{-27} \mathrm{ kg\cdot\mathrm{m/s$

Example 23: Stationary Waves in Closed and Open Pipes

A closed pipe (closed at one end, open at the other) has length $0.85 \mathrm{ m$ . Calculate the Fundamental frequency and the first two overtones. Speed of sound $= 340 \mathrm{ m/s$ .

Step 1: Fundamental frequency (first harmonic)

For a closed pipe, the fundamental has a node at the closed end and an antinode at the open end. The Effective length is $\lambda/4$ :

$f_1 = \frac{v}{4L} = \frac{340}{4 \times 0.85} = \frac{340}{3.4} = 100 \mathrm{ Hz$

Step 2: First overtone (third harmonic)

Only odd harmonics are present in a closed pipe:

$f_3 = 3f_1 = 3 \times 100 = 300 \mathrm{ Hz$

Step 3: Second overtone (fifth harmonic)

$f_5 = 5f_1 = 5 \times 100 = 500 \mathrm{ Hz$

Comparison with an open pipe of the same length:

For an open pipe: $f_1 = \frac{v}{2L} = \frac{340}{1.7} = 200 \mathrm{ Hz$ And all harmonics are Present ( $f_n = nf_1$ ).

:::info A closed pipe produces only odd harmonics (1st, 3rd, 5th, …), giving a richer, more “hollow” sound than an open pipe which produces all harmonics (1st, 2nd, 3rd, …). This is why Clarinets (effectively closed pipes) sound different from flutes (open pipes). :::

Common Pitfalls Extended

Pitfall 6: Confusing Photon Energy with Electron Energy in the Photoelectric Effect

The photon energy $E = hf$ is the energy of the incoming light. The work function $\phi$ is the Minimum energy to release an electron. The kinetic energy of the emitted electron is $KE_{\max} = hf - \phi$ . If $hf \lt \phi$ No electrons are emitted regardless of the intensity of The light.

Pitfall 7: Assuming All Transitions Are Equally Probable

Not all transitions between energy levels are equally likely. Selection rules govern which Transitions have high probability. For hydrogen, transitions where $\Delta l = \pm 1$ are strongly Favoured. However, for exam purposes, you should assume all allowed energy-conserving transitions Can occur.

Pitfall 8: Forgetting That Wave Speed Depends on the Medium

When calculating the wavelength of a wave after it enters a new medium, remember that the frequency stays the same but the wavelength and speed change:

$\lambda_2 = \frac{v_2}{v_1} \lambda_1$

This applies to all waves: light entering glass, sound entering water, waves on a string when Tension changes, etc.

Additional Practice Problems

Calculate the wavelength of (a) a $100 \mathrm{ eV$ electron and (b) a $100 \mathrm{ eV$ proton. Comment on why electron microscopes have much better resolution than optical microscopes.
A hydrogen atom is in the $n = 4$ state. Calculate the energies and wavelengths of all possible photons that could be emitted as the atom returns to the ground state. Identify which are in the visible spectrum.
In a Young’s double slit experiment using laser light of wavelength $633 \mathrm{ nm$ The slits are $0.5 \mathrm{ mm$ apart and the screen is $2 \mathrm{ m$ away. Calculate (a) the fringe spacing, (b) the distance from the central maximum to the third bright fringe, and (c) what happens to the fringe spacing if the wavelength is changed to $450 \mathrm{ nm$ .
An X-ray tube operates at $50 \mathrm{ kV$ . Calculate (a) the minimum wavelength of X-rays produced, (b) the maximum energy of the X-ray photons in eV, and (c) the momentum of these photons.
Explain how the Heisenberg uncertainty principle implies that electrons cannot exist in stationary orbits (as in the Bohr model) and must be described by probability clouds. Reference the uncertainty in both position and momentum.

Further Worked Examples

Example 26: Pair Production

A gamma ray photon converts into an electron-positron pair. Calculate the minimum photon energy Required.

Step 1: Minimum energy

The minimum photon energy must equal the rest mass energy of both particles:

$E_{\min} = 2m_e c^2 = 2 \times 0.511 = 1.022 \mathrm{ MeV$

Step 2: Corresponding wavelength

$\lambda = \frac{hc}{E} = \frac{6.626 \times 10^{-34} \times 3.0 \times 10^8}{1.022 \times 10^6 \times 1.602 \times 10^{-19}}$

$= \frac{1.988 \times 10^{-25}}{1.637 \times 10^{-13}} = 1.214 \times 10^{-12} \mathrm{ m$

This is in the gamma ray region of the electromagnetic spectrum.

:::info Pair production cannot occur in a vacuum — it must happen near a nucleus to conserve Momentum. The recoil of the nucleus absorbs the excess momentum. This is why pair production is more Likely in high-atomic-number materials (lead, tungsten), which are used in radiation shielding. :::

Example 27: Energy Levels of Hydrogen — Detailed Transitions

Calculate the wavelength and frequency of the photon emitted when a hydrogen atom transitions from $n = 4$ to $n = 2$ (H-beta line in the Balmer series).

Step 1: Energy of each level

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

$E_4 = -\frac{13.6}{16} = -0.85 \mathrm{ eV$

$E_2 = -\frac{13.6}{4} = -3.40 \mathrm{ eV$

Step 2: Energy of emitted photon

$\Delta E = E_4 - E_2 = -0.85 - (-3.40) = 2.55 \mathrm{ eV$

Step 3: Wavelength

$\lambda = \frac{hc}{\Delta E} = \frac{1240 \mathrm{ eV\cdot\mathrm{nm}{2.55 \mathrm{ eV} = 486.3 \mathrm{ nm$

This is in the blue-green region of the visible spectrum — the H-beta line.

Step 4: Frequency

$f = \frac{c}{\lambda} = \frac{3.0 \times 10^8}{4.863 \times 10^{-7}} = 6.17 \times 10^{14} \mathrm{ Hz$

Example 28: Standing Wave on a String — All Harmonics

A string of length $0.75 \mathrm{ m$ and mass $0.015 \mathrm{ kg$ is under a tension of $40 \mathrm{ N$ . Calculate the fundamental frequency, the first three harmonic frequencies, and the Wave speed.

Step 1: Wave speed

$v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{40}{0.015/0.75}} = \sqrt{\frac{40}{0.02}} = \sqrt{2000} = 44.7 \mathrm{ m/s$

Step 2: Fundamental frequency

$f_1 = \frac{v}{2L} = \frac{44.7}{1.5} = 29.8 \mathrm{ Hz$

Step 3: Harmonics

$f_2 = 2f_1 = 59.6 \mathrm{ Hz$

$f_3 = 3f_1 = 89.4 \mathrm{ Hz$

$f_4 = 4f_1 = 119.2 \mathrm{ Hz$

All harmonics are present for a string fixed at both ends.

Board-Specific Content: SQA Advanced Higher

Wave-Particle Duality: Davisson-Germer Experiment

The Davisson-Germer experiment (1927) provided direct evidence for the wave nature of electrons. Electrons were accelerated through a potential difference and directed at a nickel crystal. The Reflected electrons showed a diffraction pattern, with a peak at an angle consistent with Bragg’s Law:

$n\lambda = 2d\sin\theta$

For electrons accelerated through $54 \mathrm{ V$ :

$\lambda = \frac{1.226}{\sqrt{54}} = 0.167 \mathrm{ nm$

The observed peak angle gave $\lambda = 0.165 \mathrm{ nm$ In excellent agreement. This confirmed de Broglie’s hypothesis that $\lambda = h/p$ .

Standard Model Particles

The Standard Model classifies fundamental particles into:

Quarks (6 flavours: up, down, charm, strange, top, bottom) — experience all four forces
Leptons (6: electron, muon, tau + their neutrinos) — experience gravity, EM, weak
Gauge bosons (force carriers): photon (EM), gluon (strong), W and Z (weak), Higgs (mass)
Each quark comes in three colours (red, green, blue) — the charge of the strong force

Conservation Laws in Particle Physics

In every particle interaction, the following must be conserved:

Energy/momentum
Electric charge
Baryon number
Lepton number (electron lepton number, muon lepton number, tau lepton number separately)
Strangeness (in strong interactions only; can change by 1 in weak interactions)

Additional Practice Problems

Calculate the energy of a photon with wavelength $0.02 \mathrm{ nm$ (hard X-ray region). Can this photon produce pair production? Justify your answer.
An electron in a hydrogen atom is in the $n = 5$ state. List all possible transitions that emit photons in the visible spectrum and calculate their wavelengths.
A string under tension $T$ has fundamental frequency $f_1$ . By what factor must the tension change to double the fundamental frequency? By what factor must the length change?
A positron with kinetic energy $2 \mathrm{ MeV$ collides with an electron at rest, producing two gamma ray photons. Calculate the wavelength of each photon (assume they have equal energy).
Compare and contrast the properties of alpha, beta, and gamma radiation in a table. Include: nature, charge, mass, speed, penetrating power, ionising power, and deflection in magnetic fields.

Mark this page as reviewed