Chapter 11: Dual Nature of Radiation and Matter

The early 20th century revealed that light, long thought of as a wave (Maxwell, Young), behaves as a stream of particles (photons) in the photoelectric effect. Conversely, electrons, supposedly classical particles, were found to diffract like waves (Davisson-Germer). This wave-particle duality is the hallmark of quantum mechanics.

Concept Map

• Electron emission $\rightarrow$ thermionic / photoelectric / field / secondary.
• Photoelectric effect $\rightarrow$ classical failures $\rightarrow$ Einstein's photon hypothesis $\rightarrow$ $KE_{\max} = h\nu - \phi$.
• Photons $\rightarrow$ $E = h\nu$, $p = h/\lambda$, rest mass zero.
• Matter waves $\rightarrow$ de Broglie $\lambda = h/p$ $\rightarrow$ electron through $V$: $\lambda = 12.27/\sqrt V$ Å.
• Davisson-Germer $\rightarrow$ electron diffraction confirms duality.
• Uncertainty principle $\rightarrow$ $\Delta x \Delta p \geq \hbar/2$.

11.1 Electron Emission and Work Function

Definition

Electrons in a metal are held by the binding energy of the lattice. The work function $\phi$ is the minimum energy required to free an electron from the surface. Typical values: Cs $\sim 2.14\,\text{eV}$, Na $\sim 2.75$, Cu $\sim 4.65$, Pt $\sim 5.65$.

Four mechanisms supply this energy:

Pitfalls

• $\phi$ is a property of the surface, not the bulk; oxide layers can change it.
• "Work function in eV" is convenient; converting to joules requires the factor $1\,\text{eV} = 1.6 \times 10^{-19}\,\text{J}$.

11.2 Photoelectric Effect — Observations

Discovery

Hertz (1887) noticed that UV light shining on a spark-gap electrode made discharges easier. Lenard (1900) showed that the ejected particles were negatively charged with the same $e/m$ as cathode rays — they were electrons.

Setup

An evacuated tube contains a photosensitive cathode $C$ and a collector anode $A$. Light of frequency $\nu$ illuminates $C$; emitted photoelectrons reach $A$, giving a photocurrent $I$ measured by a galvanometer. A variable voltage $V$ (with reversible polarity) accelerates or retards the electrons.

Pitfalls

• Photocurrent is observed only above a threshold frequency — even very intense low-frequency light cannot eject any electron.
• The effect is essentially instantaneous ($< 10^{-9}\,\text{s}$), contradicting wave-theory predictions of long lag times.

11.3 Effect of Intensity, Potential, Frequency

Photocurrent vs Anode Potential (fixed $\nu$, varying intensity)

Plot photocurrent $I$ vs anode potential $V$:

• For positive $V$, the photocurrent rises and saturates at a value $I_{\text{sat}}$ that depends on intensity.
• Saturation $I_{\text{sat}} \propto$ intensity: more photons $\to$ more photoelectrons per second.
• For negative $V$ (retarding), photocurrent falls. The minimum negative $V$ at which the photocurrent becomes zero is the stopping potential $V_0$ — and it is independent of intensity.

Stopping potential vs Frequency

For a given metal, varying $\nu$ (at fixed intensity):

• Below threshold $\nu_0$: no current at any intensity.
• Above $\nu_0$: $V_0$ increases linearly with $\nu$. Slope $= h/e$ (universal, independent of metal); intercept on the $\nu$-axis $= \nu_0$ (metal-dependent).

Graph summary

Pitfalls

• Stopping potential is independent of intensity but depends on frequency and metal.
• Saturation current depends on intensity but not on $V$ once $V$ is large enough.

11.4 Threshold Frequency and Classical Wave-Theory Failures

Classical predictions (Maxwell)

1. Higher intensity $\to$ larger $E$ field $\to$ more KE per electron.
2. No threshold frequency: even low-$\nu$ light should eventually eject electrons given enough time.
3. Time lag: for weak intensity, light's energy must accumulate at an atom — measurable lag time of $\sim$ minutes/hours predicted.

Experimental facts

1. KE of photoelectrons is independent of intensity. Higher intensity only increases the number of photoelectrons.
2. Below threshold $\nu_0$, no current is observed at any intensity.
3. The effect is virtually instantaneous ($< 10^{-9}\,\text{s}$), even at very low intensity.

These three observations falsify the wave picture for the energy-transfer step.

Pitfalls

• The wave model is fine for diffraction and interference; it fails specifically for single-photon-absorption energetics.

11.5 Einstein's Photoelectric Equation

Einstein's hypothesis (1905)

Light of frequency $\nu$ is delivered in discrete packets — photons — each carrying energy

$E = h\nu.$

When a photon is absorbed by an electron in the metal, the electron uses up to $\phi$ of that energy to escape and keeps the rest as kinetic energy. Hence

$\boxed{\;KE_{\max} = h\nu - \phi\;}.$

If $\nu < \nu_0 \equiv \phi/h$, no electron has enough energy to escape — explains threshold.

Stopping potential

The most energetic electrons are stopped when $eV_0 = KE_{\max}$:

$eV_0 = h\nu - \phi \quad\Rightarrow\quad V_0 = \frac{h}{e}\nu - \frac{\phi}{e}.$

So $V_0$ vs $\nu$ is linear with slope $h/e$ and $x$-intercept $\nu_0$.

Explanation of all observations

• Intensity $\propto$ number of photons / sec $\propto$ number of ejected electrons / sec $\propto I_{\text{sat}}$. KE per electron unchanged.
• Threshold $\nu_0 = \phi/h$ arises naturally.
• One photon ejects one electron in a single quantum event — explains instantaneous response.

Worked Example

Light of $\lambda = 400\,\text{nm}$ falls on Cs ($\phi = 2.14\,\text{eV}$). Find $V_0$.

$h\nu = hc/\lambda = (1240\,\text{eV nm})/400 = 3.1\,\text{eV}$. $KE_{\max} = 3.1 - 2.14 = 0.96\,\text{eV}$. $V_0 = 0.96\,\text{V}$.

Pitfalls

• $\phi$ is in joules or eV — match units with $h\nu$.
• The "$\max$" in $KE_{\max}$ matters: most electrons emerge with less because they lose energy reaching the surface; only the surface-layer ones have $KE_{\max}$.
• $eV_0 = KE_{\max}$ is the only condition tested — applies to maximum KE, not average.

11.6 Photon — Properties

A photon, the quantum of the EM field, has

• Energy $E = h\nu = hc/\lambda$.
• Momentum $p = E/c = h/\lambda$.
• Rest mass zero (always travels at $c$).
• Spin $1$ (boson).
• Particle-like in interactions (photoelectric effect, Compton scattering); wave-like in propagation (interference, diffraction).

Number of photons per second

A source of power $P$ at frequency $\nu$ emits

$N = \frac{P}{h\nu}\quad\text{photons / s}.$

Radiation pressure

A photon absorbed by a surface delivers momentum $p = h/\lambda$. For light of intensity $I$ on a perfectly absorbing surface:

$P_{\text{rad}} = \frac{I}{c}.$

For a perfectly reflecting surface: $P_{\text{rad}} = 2I/c$.

Worked Example

$\lambda = 500\,\text{nm}$. Find $E$ and $p$ of a photon.

$E = hc/\lambda = (6.63\times 10^{-34})(3\times 10^8)/(5\times 10^{-7}) = 3.98\times 10^{-19}\,\text{J} = 2.48\,\text{eV}$. $p = h/\lambda = (6.63\times 10^{-34})/(5\times 10^{-7}) = 1.33\times 10^{-27}\,\text{kg m/s}$.

Pitfalls

• Photon mass $m = 0$, but momentum $p = h/\lambda$ is nonzero — relativistically, $E = pc$ requires $m_0 = 0$.
• A photon's frequency is the same in any inertial frame's measurement of the same event count, but Doppler effect changes $\nu$ between source and observer frames.

11.7 de Broglie Hypothesis

Hypothesis (1924)

If radiation has dual nature, so does matter. A particle of momentum $p$ is associated with a wave of wavelength

$\boxed{\;\lambda = \frac{h}{p} = \frac{h}{mv}\;}.$

For non-relativistic particles, $p = mv$; for relativistic, $p = \gamma m v$.

For a particle of kinetic energy $K$ (non-relativistic): $p = \sqrt{2mK}$, so

$\lambda = \frac{h}{\sqrt{2mK}}.$

Worked Example

A bullet of mass $10\,\text{g}$ moves at $1000\,\text{m/s}$. Find $\lambda$.

$\lambda = (6.63\times 10^{-34})/(0.01\cdot 1000) = 6.63\times 10^{-35}\,\text{m}$. Far below any atomic scale, hence not observable as wave.

For an electron of $K = 1\,\text{eV}$: $\lambda = h/\sqrt{2mK} \approx 1.23\,\text{nm}$. Observable by diffraction in a crystal lattice.

Pitfalls

• de Broglie's $\lambda$ is meaningful only when $\lambda$ is comparable to the system's relevant size (slit width, lattice spacing, atom size).
• For macroscopic objects, $\lambda$ is absurdly small — quantum effects negligible.

11.8 de Broglie Wavelength of an Electron in Potential $V$

Derivation: $\lambda = 12.27/\sqrt V$ Å

An electron accelerated through potential $V$ gains kinetic energy $K = eV$. Then

$p = \sqrt{2 m_e e V}, \quad \lambda = \frac{h}{\sqrt{2 m_e e V}}.$

Plug numbers: $h = 6.626\times 10^{-34}$ J s, $m_e = 9.11\times 10^{-31}$ kg, $e = 1.6\times 10^{-19}$ C:

$\lambda = \frac{6.626\times 10^{-34}}{\sqrt{2\cdot 9.11\times 10^{-31}\cdot 1.6\times 10^{-19}\cdot V}}.$

The denominator $= 5.40\times 10^{-25}\sqrt V$ (SI), so $\lambda = (1.227\times 10^{-9}/\sqrt V)\,\text{m} = (12.27/\sqrt V)$ Å.

$\boxed{\;\lambda_e = \frac{12.27}{\sqrt V}\;\text{Å},\quad V\text{ in volts}\;}.$

Worked Example

Electron accelerated through $V = 100\,\text{V}$.

$\lambda = 12.27/10 = 1.227\,\text{Å}$. Comparable to atomic spacing; perfect for diffraction off a crystal.

Pitfalls

• Use $V$ in volts, get $\lambda$ in Å. For $V$ in kV, you must scale: $\lambda = 12.27/\sqrt{V(\text{V})}$ Å $= 0.388/\sqrt{V(\text{kV})}$ Å.
• For electron $V \gtrsim 100\,\text{kV}$, relativistic correction is needed.

11.9 Davisson-Germer Experiment

Setup

A heated tungsten filament emits electrons, which are accelerated by a variable potential $V$ and directed normally onto a polished single crystal of nickel. A movable Faraday-cup detector measures the intensity of scattered electrons as a function of scattering angle $\phi$ at fixed $V$.

Observation

For $V = 54\,\text{V}$, a sharp maximum in scattered intensity is found at $\phi = 50^\circ$ — a "diffraction peak" from the nickel lattice.

Analysis

Treating the Ni atomic rows as a diffraction grating with spacing $d = 0.91\,\text{Å}$ (for Ni $(111)$ plane), Bragg-like condition gives wavelength of the scattering wave: $\lambda = d\sin\phi = 0.91\sin 50^\circ \approx 0.165\,\text{nm} = 1.65\,\text{Å}$ (using the appropriate plane and incidence). Cross-checking with de Broglie: $\lambda = 12.27/\sqrt{54} = 1.67\,\text{Å}$.

Conclusion

The measured wavelength matches the de Broglie wavelength of the electrons. Electrons therefore behave as waves; matter waves are real.

Pitfalls

• The "peak angle" depends on $V$; varying $V$ shifts the peak in agreement with $\lambda \propto 1/\sqrt V$.
• This was an accidental discovery — Davisson and Germer were originally studying electron scattering from oxidised Ni; an accident annealed the Ni into a single crystal.

11.10 Heisenberg Uncertainty Principle (Qualitative)

Statement

For any quantum particle, one cannot simultaneously know position $x$ and momentum $p_x$ with arbitrary precision:

$\Delta x\,\Delta p_x \geq \frac{\hbar}{2}, \quad \hbar = \frac{h}{2\pi}.$

Analogous: $\Delta E\,\Delta t \geq \hbar/2$.

Intuition from de Broglie

If a particle is localised in space within $\Delta x$, it must be built from a wave packet — a superposition of many de Broglie waves with momentum spread $\Delta p \sim h/\Delta x$.

Implications

• The classical orbit of an electron in an atom has no well-defined trajectory.
• Zero-point energy: a confined particle (in a box of size $L$) has minimum KE $\sim \hbar^2/(2mL^2)$.
• Cannot "see" an electron with light shorter than its size without imparting huge momentum to it.

Worked Example

An electron is confined to an atom of size $\Delta x = 1\,\text{Å}$. Estimate $\Delta p$ and corresponding KE.

$\Delta p \geq \hbar/(2\Delta x) = (1.05\times 10^{-34})/(2\cdot 10^{-10}) = 5.3\times 10^{-25}\,\text{kg m/s}$.

$KE \sim (\Delta p)^2/(2m_e) = (5.3\times 10^{-25})^2/(2\cdot 9.11\times 10^{-31}) \approx 1.5\times 10^{-19}\,\text{J} \approx 1\,\text{eV}$ — sensible atomic scale.

Pitfalls

• $\Delta x$ and $\Delta p$ are standard deviations, not measurement errors.
• The uncertainty principle is not about disturbing the system in measurement; it is an intrinsic feature of the quantum state.

Solved Problems

1. Light of $\lambda = 200\,\text{nm}$ hits a metal of $\phi = 4.0\,\text{eV}$. Find $V_0$.

$h\nu = 1240/200 = 6.2\,\text{eV}$. $V_0 = (6.2 - 4.0) = 2.2\,\text{V}$.

2. A metal has threshold $\lambda_0 = 600\,\text{nm}$. Find $\phi$.

$\phi = hc/\lambda_0 = 1240/600 = 2.07\,\text{eV}$.

3. Slope of $V_0$ vs $\nu$ is found to be $4.12\times 10^{-15}\,\text{V s}$. Find $h$.

$h = e\cdot\text{slope} = (1.6\times 10^{-19})(4.12\times 10^{-15}) = 6.59\times 10^{-34}\,\text{J s}$. (Standard $h = 6.626\times 10^{-34}\,\text{J s}$.)

4. Power $5\,\text{mW}$ laser at $\lambda = 633\,\text{nm}$. Photon rate?

$E_{\text{photon}} = 1240/633 = 1.96\,\text{eV} = 3.14\times 10^{-19}\,\text{J}$. $N = P/E = 5\times 10^{-3}/3.14\times 10^{-19} \approx 1.6\times 10^{16}$ photons/s.

5. A $1\,\text{eV}$ electron's de Broglie wavelength?

$\lambda = 12.27/\sqrt 1 = 12.27\,\text{Å}$.

6. Find the ratio $\lambda_{\text{e}}/\lambda_{\text{p}}$ for an electron and a proton of equal KE.

$\lambda \propto 1/\sqrt m$ at equal $K$, so $\lambda_e/\lambda_p = \sqrt{m_p/m_e} = \sqrt{1836} \approx 42.85$.

7. A particle of $\Delta x = 10^{-10}\,\text{m}$. Minimum $\Delta v$ of an electron in it?

$\Delta v = \hbar/(2 m_e \Delta x) = (1.05\times 10^{-34})/(2\cdot 9.11\times 10^{-31}\cdot 10^{-10}) \approx 5.8\times 10^5\,\text{m/s}$.

JEE/NEET Edge Cases

• A common trap: "doubling intensity doubles $V_0$" — false. $V_0$ depends only on $\nu$ and $\phi$.
• Threshold wavelength $\lambda_0$ and frequency $\nu_0$: $\lambda_0\nu_0 = c$; $\phi = h\nu_0 = hc/\lambda_0$.
• For a photon, $p = h/\lambda = E/c$; for matter, $p = h/\lambda$ but $E \neq pc$ (because matter has rest mass).
• For equal momenta $\lambda_e = \lambda_p$; for equal KE $\lambda \propto 1/\sqrt m$; for equal velocity $\lambda \propto 1/m$.
• de Broglie wavelength of a neutron in a thermal reactor ($K \approx 0.025\,\text{eV}$): $\lambda \approx 1.8\,\text{Å}$ — perfect for crystallography.
• The $V_0$-vs-$\nu$ line is the same slope for all metals (it's $h/e$, universal); only the threshold $\nu_0$ depends on the metal.
• Saturating current $\propto$ intensity for fixed $\nu$. Reducing intensity to almost zero still gives instantaneous photoelectrons above threshold — one photon, one electron.
• Davisson-Germer specifically supports the wave nature of electrons; G P Thomson's transmission diffraction through gold foil is its other classic confirmation.

Quick Recap

• Work function $\phi$: minimum energy to free an electron. $\nu_0 = \phi/h$.
• Einstein: $KE_{\max} = h\nu - \phi$; $eV_0 = h\nu - \phi$.
• $V_0$ vs $\nu$ linear, slope $h/e$.
• Photon: $E = h\nu$, $p = h/\lambda$, $m_0 = 0$.
• de Broglie: $\lambda = h/p = h/\sqrt{2mK}$. Electron: $\lambda = 12.27/\sqrt V$ Å.
• Davisson-Germer: $54\,\text{V}$ electron diffracts off Ni at $50^\circ$, $\lambda \approx 1.65\,\text{Å}$.
• Heisenberg: $\Delta x \Delta p \geq \hbar/2$.

Formula Sheet