Dual Nature of Light and Matter
(Physics – KCET Study Notes)
1. Simple Explanation
- Dual nature means that light and matter can behave both as waves and as particles depending on the experiment.
- In a wave picture we use concepts such as wavelength, frequency, interference, and diffraction.
- In a particle picture we use photons (for light) or electrons/atoms (for matter) that carry discrete packets of energy/momentum.
This duality is a cornerstone of quantum mechanics and explains many phenomena that classical physics cannot.
2. Key Points
| Aspect | Wave Behaviour | Particle Behaviour |
|---|---|---|
| Light | Interference (double‑slit), diffraction (gratings), polarisation | Photoelectric effect (energy (E = h\nu)), Compton scattering (wavelength shift) |
| Matter | Electron diffraction (Davisson–Germer), de Broglie wavelength (\lambda = h/p) | Electron impact on crystals, quantised energy levels in atoms |
| Key Experiments | 1. Young’s double‑slit (light) 2. Davisson–Germer (electrons) 3. Compton (X‑rays) |
1. Photoelectric effect (Heaviside, Einstein) 2. Bohr model (quantised orbits) |
| Important Constants | (h) – Planck’s constant (6.626\times10^{-34},J,s) (c) – speed of light (3.00\times10^8,m/s) |
(\hbar = h/2\pi) |
| Relations | (E = h\nu = \frac{hc}{\lambda}) (light) (\lambda = \frac{h}{p}) (matter) |
(p = mv) (classical) (E = \frac{p^2}{2m}) (kinetic) |
| Wave‑Particle Duality | Wave explains interference & diffraction; particle explains localized impacts & energy quanta. | Particle explanation of photoelectric effect; wave explanation of electron diffraction. |
3. Worked Example
Problem
An electron is accelerated through a potential difference of (400,V).
- Find the kinetic energy (in eV).
- Calculate the de Broglie wavelength (in picometres, pm).
- Determine the diffraction angle for the first order ((m=1)) when it strikes a crystal with lattice spacing (d = 0.20,nm).
Step‑by‑Step Solution
1. Kinetic Energy
When an electron is accelerated through a potential (V), the gained kinetic energy (K) equals the electric potential energy: [ K = eV ] where (e = 1.602\times10^{-19},C) is the elementary charge.
[ K = (1.602\times10^{-19},C)(400,V) = 6.408\times10^{-17},J ]
Convert to electron‑volts: [ K = 400,eV ] (Directly (eV) units: 1 eV is the energy gained by one electron through 1 V potential.)
2. de Broglie Wavelength
[ \lambda = \frac{h}{p} = \frac{h}{mv} ]
First find the speed (v) from kinetic energy: [ K = \frac{1}{2}mv^2 ;\Rightarrow; v = \sqrt{\frac{2K}{m}} ] Electron rest mass (m = 9.11\times10^{-31},kg).
[ v = \sqrt{\frac{2\times6.408\times10^{-17}}{9.11\times10^{-31}}} = \sqrt{1.406\times10^{15}} \approx 1.186\times10^{7},m/s ]
Now compute momentum (p = mv): [ p = (9.11\times10^{-31})(1.186\times10^{7}) \approx 1.081\times10^{-23},kg,m/s ]
Finally, [ \lambda = \frac{6.626\times10^{-34}}{1.081\times10^{-23}} \approx 6.13\times10^{-11},m = 61.3,pm ]
3. Diffraction Angle (Bragg’s Law)
For an electron beam diffracting from crystal planes with spacing (d), the Bragg condition is: [ n\lambda = 2d\sin\theta ] with (n = 1) for first order.
[ \sin\theta = \frac{\lambda}{2d} = \frac{61.3\times10^{-12}}{2\times0.20\times10^{-9}} = \frac{61.3}{400}\times10^{-3} \approx 0.153 ]
[ \theta = \sin^{-1}(0.153) \approx 8.8^{\circ} ]
Answer
- Kinetic energy = 400 eV
- de Broglie wavelength ≈ 61 pm
- First‑order diffraction angle ≈ 8.8°
4. Quick Revision Summary
- Dual nature: Light and matter act as waves and particles.
- Wave signatures: Interference, diffraction, polarisation, de Broglie wavelength.
- Particle signatures: Photoelectric effect, Compton scattering, quantised