Physics Dual Nature of Matter and Radiation

Dual Nature of Matter and Radiation Formula Sheet

All key Dual Nature formulas: photoelectric effect, photon energy/momentum, de Broglie wavelength, Davisson-Germer Bragg law. JEE Main & Advanced quick revision.

6 min read Updated Jun 2026 #formula sheet#quick revision#jee-main

Last-minute revision sheet for the entire Dual Nature chapter: photoelectric effect, photon theory, de Broglie matter waves, and the Davisson-Germer experiment. Every formula, constant, and shortcut below is pulled straight from the chapter pages.

Chapter Map

graph TD
    A[Dual Nature] --> B[Photoelectric Effect]
    A --> C[Photon Theory]
    A --> D[Matter Waves / de Broglie]
    A --> E[Davisson-Germer]
    B --> B1[Einstein's Equation]
    B --> B2[Stopping Potential]
    C --> C1[Energy & Momentum]
    C --> C2[Radiation Pressure]
    D --> D1[de Broglie Wavelength]
    D --> D2[Bohr Quantization]
    E --> E1[Bragg's Law]

Photoelectric Effect

Einstein’s photoelectric equation is the headline result of this chapter.

$$\boxed{K_{max} = h\nu - \phi}$$$$\frac{1}{2}mv_{max}^2 = h\nu - \phi$$

In terms of wavelength:

$$K_{max} = \frac{hc}{\lambda} - \phi$$

Work Function and Threshold

$$\boxed{\phi = h\nu_0 = \frac{hc}{\lambda_0}}$$$$\lambda_0 = \frac{hc}{\phi}$$

At threshold, $K_{max} = 0$, so $h\nu_0 = \phi$.

Stopping Potential

$$\boxed{eV_0 = K_{max} = h\nu - \phi}$$$$V_0 = \frac{h}{e}\nu - \frac{\phi}{e} = \frac{h}{e}(\nu - \nu_0)$$

This is a straight line $y = mx + c$ with slope $= h/e$ and y-intercept $= -\phi/e$.

QuantityFormulaNotes
Photon energy$E = h\nu = \dfrac{hc}{\lambda}$Energy of one incident photon
Max KE of photoelectron$K_{max} = h\nu - \phi$Depends only on frequency
Work function$\phi = h\nu_0 = \dfrac{hc}{\lambda_0}$Minimum energy to free an electron
Stopping potential$eV_0 = K_{max}$Independent of intensity
$V_0$ vs $\nu$ slope$\dfrac{h}{e}$Universal constant $= 4.14 \times 10^{-15}$ V·s
Threshold wavelength$\lambda_0 = \dfrac{hc}{\phi}$Longest $\lambda$ that ejects electrons

Comparison Relations

Ratio of stopping potentials for two frequencies:

$$\frac{V_{01}}{V_{02}} = \frac{h\nu_1 - \phi}{h\nu_2 - \phi}$$

Change in stopping potential:

$$\Delta V_0 = \frac{h}{e}\Delta\nu = \frac{hc}{e}\left(\frac{1}{\lambda_1} - \frac{1}{\lambda_2}\right)$$
High-Yield: Frequency vs Intensity

Frequency controls $K_{max}$ and stopping potential. Intensity controls the number of photoelectrons (photocurrent). Doubling intensity doubles photocurrent but leaves $K_{max}$ and $V_0$ unchanged. Emission is instantaneous ($\sim 10^{-9}$ s), with no time lag.

Key Experimental Observations

  1. Below threshold frequency $\nu_0$: no emission, regardless of intensity.
  2. Saturation current is proportional to intensity, independent of frequency.
  3. Stopping potential $V_0 \propto \nu$ (linear), independent of intensity.
  4. Emission is instantaneous (no time lag).

Photon Theory

PropertyFormulaNotes
Energy$E = h\nu = \dfrac{hc}{\lambda}$Discrete energy packet
Momentum$p = \dfrac{h\nu}{c} = \dfrac{h}{\lambda}$Photon carries momentum
Rest mass$0$Photons always travel at $c$
Relativistic mass$m = \dfrac{h\nu}{c^2} = \dfrac{h}{\lambda c}$Effective mass equivalent

Radiation Pressure

$$\boxed{P = \frac{I}{c} \text{ (complete absorption)}, \qquad P = \frac{2I}{c} \text{ (complete reflection)}}$$

where $I$ is the intensity of incident radiation.

de Broglie Hypothesis

Every moving particle has an associated matter wave.

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

All Forms of the de Broglie Wavelength

Known quantityFormulaNotes
Velocity $v$$\lambda = \dfrac{h}{mv}$Free particle
Kinetic energy $K$$\lambda = \dfrac{h}{\sqrt{2mK}}$Using $p = \sqrt{2mK}$
Accelerating potential $V$, charge $q$$\lambda = \dfrac{h}{\sqrt{2mqV}}$$K = qV$
Electron through $V$ volts$\lambda = \dfrac{12.27}{\sqrt{V}}$ ÅMemorize this shortcut
Electron with KE $K$ (eV)$\lambda = \dfrac{12.27}{\sqrt{K}}$ Å$V$ replaced by $K$ in eV
Thermal neutron at $T$$\lambda = \dfrac{h}{\sqrt{3mkT}}$Uses $K = \tfrac{3}{2}kT$

Relating momentum and kinetic energy:

$$K = \frac{1}{2}mv^2 = \frac{p^2}{2m} \quad\Rightarrow\quad p = \sqrt{2mK}$$

Electron-specific shortcut:

$$\boxed{\lambda = \frac{h}{\sqrt{2meV}} = \frac{12.27}{\sqrt{V}} \text{ Å}}$$

Relativistic Form

For high speeds ($v \sim c$):

$$\lambda = \frac{h}{\sqrt{2m_0 K\left(1 + \dfrac{K}{2m_0 c^2}\right)}}$$

Phase and Group Velocity

VelocityFormulaMeaning
Phase velocity$v_p = \nu\lambda = \dfrac{\omega}{k} = \dfrac{E}{p} = \dfrac{c^2}{v}$$v_p > c$, carries no information
Group velocity$v_g = \dfrac{d\omega}{dk} = v$Actual particle velocity
Product$v_p \cdot v_g = c^2$Key relation
Inverse Proportionalities
$\lambda \propto \dfrac{1}{v}$, $\quad \lambda \propto \dfrac{1}{\sqrt{K}}$, $\quad \lambda \propto \dfrac{1}{\sqrt{V}}$. At the same kinetic energy: $\lambda \propto \dfrac{1}{\sqrt{m}}$ (lighter particle has larger $\lambda$). At the same accelerating potential: $\lambda \propto \dfrac{1}{\sqrt{mq}}$ (used to compare different particles). Heavier or faster particles have smaller wavelengths and behave more particle-like.

Bohr Quantization from de Broglie

A stable orbit fits a whole number of wavelengths (standing wave):

$$2\pi r = n\lambda$$

Substituting $\lambda = h/mv$:

$$\boxed{mvr = \frac{nh}{2\pi} = n\hbar}$$

This is exactly Bohr’s angular momentum quantization condition.

Photon vs Electron

PropertyPhotonElectron
Rest mass$0$$9.1 \times 10^{-31}$ kg
Charge$0$$-1.6 \times 10^{-19}$ C
Speed$c$ always$v < c$
Wavelength$\dfrac{hc}{E}$$\dfrac{h}{mv}$

For a photon, de Broglie wavelength equals the electromagnetic wavelength: $E = pc \Rightarrow p = h/\lambda_{EM}$, so $\lambda_{dB} = \lambda_{EM}$.

Davisson-Germer Experiment

Confirmed the wave nature of electrons via diffraction from a nickel single crystal (1927).

Bragg’s Law

$$\boxed{2d\sin\theta = n\lambda}$$

where $d$ = crystal plane spacing, $\theta$ = glancing angle, $n$ = order (usually 1), $\lambda$ = wavelength.

The chapter overview also states the diffraction condition as $d\sin\theta = n\lambda$; in the detailed Davisson-Germer analysis the working relation is the Bragg form $2d\sin\theta = n\lambda$.

Angle Conversion

$$\theta = \frac{180^\circ - \varphi}{2}$$

where $\varphi$ is the scattering angle and $\theta$ is the glancing angle (the angle used in Bragg’s law).

The Classic 54 V Result

QuantityValueNotes
Accelerating voltage $V$54 V
Scattering angle $\varphi$50°Observed peak
Glancing angle $\theta$65°$\theta = (180^\circ - 50^\circ)/2$
Crystal spacing $d$ (Ni, 111)0.91 ÅFCC nickel
$\lambda$ from de Broglie$\dfrac{12.27}{\sqrt{54}} = 1.67$ Å
$\lambda$ from Bragg’s law$2d\sin\theta \approx 1.65$ ÅAgreement within ~1%
High-Yield: Davisson-Germer Pitfalls

Always use the glancing angle $\theta$ in Bragg’s law, never the scattering angle $\varphi$. Remember $\lambda = 12.27/\sqrt{V}$ (inverse, not $\times\sqrt{V}$). For first-order diffraction to occur you need $\lambda \le 2d$. The magic numbers to recall: 54 V, 50°, 65°, 1.67 Å.

Heisenberg’s Uncertainty Principle

$$\boxed{\Delta x \cdot \Delta p \geq \frac{h}{4\pi}}$$$$\Delta E \cdot \Delta t \geq \frac{h}{4\pi}$$

Implications: position and momentum cannot be known simultaneously with arbitrary precision; this explains why electrons do not collapse into the nucleus.

Constants and Quick Values

ConstantSymbolValue
Planck’s constant$h$$6.626 \times 10^{-34}$ J·s
Planck’s constant$h$$4.14 \times 10^{-15}$ eV·s
$h/e$ (slope of $V_0$ vs $\nu$)$4.14 \times 10^{-15}$ V·s
Golden constant$hc$$1240$ eV·nm
Electron charge$e$$1.6 \times 10^{-19}$ C
Electron mass$m_e$$9.11 \times 10^{-31}$ kg
Proton mass$m_p$$1.67 \times 10^{-27}$ kg
Speed of light$c$$3 \times 10^8$ m/s
Proton-to-electron mass ratio$m_p/m_e$$\approx 1836$
Use hc = 1240 eV·nm

For photoelectric problems, $E(\text{eV}) = \dfrac{1240}{\lambda(\text{nm})}$ saves time. For electron matter-wave problems, jump straight to $\lambda(\text{Å}) = \dfrac{12.27}{\sqrt{V}}$ with $V$ in volts.

Work Functions of Common Metals

Metal$\phi$ (eV)$\nu_0$ ($10^{14}$ Hz)$\lambda_0$ (nm)
Cesium (Cs)2.15.1590
Sodium (Na)2.35.6540
Potassium (K)2.35.6540
Calcium (Ca)2.97.0430
Zinc (Zn)4.310.4290
Silver (Ag)4.711.4260
Platinum (Pt)5.613.5220

Work function ladder (easy to hard to eject): Cs < Na < K < Al < Cu < Zn < Ag < Pt.

Common Mistakes to Avoid

WrongRight
$K_{max} = h\lambda - \phi$$K_{max} = \dfrac{hc}{\lambda} - \phi$ (inverse in $\lambda$)
Intensity determines emissionFrequency must exceed $\nu_0$; intensity sets electron count
All photoelectrons have $K_{max}$Electrons have $0 \le K \le K_{max}$
$\lambda = h/\sqrt{mK}$$\lambda = h/\sqrt{2mK}$ (don’t drop the 2)
$\lambda = 12.27\sqrt{V}$$\lambda = 12.27/\sqrt{V}$ Å
Use scattering angle $\varphi$ in Bragg’s lawUse glancing angle $\theta = (180^\circ - \varphi)/2$
Particle travels at $v_p = c^2/v$Particle travels at $v_g = v$