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Dual nature of matter and radiation is one of those JEE topics where one concept beautifully links quantum physics with classical ideas. Expect direct formula-based numericals as well as conceptual MCQs.
Dual Nature of Matter and Radiation belongs to Modern Physics and typically carries 2–3 questions (8–12 marks) in JEE Main and at least one question in JEE Advanced almost every year.
Dual Nature of Matter and Radiation JEE Notes
The chapter explains why light sometimes behaves like particles (photons) and why material particles sometimes behave like waves (de Broglie waves). You will learn how Einstein’s photoelectric equation quantifies photon energy and how de Broglie wavelength governs electron diffraction. These JEE Physics Notes discuss:
- Experimental evidence for dual behaviour: Photoelectric effect, electron diffraction.
- Key equations linking energy, momentum and wavelength.
- Graphical analysis used in typical MCQs.
- Common traps and quick tips to save time.
Photoelectric Effect and Photons
1. Experimental Observations
- No photoelectrons are emitted below a certain frequency (threshold frequency $$\nu_0$$) irrespective of intensity.
- Above $$\nu_0$$, photoelectric current appears without any time lag.
- Maximum kinetic energy of emitted electrons depends on frequency, not on intensity.
2. Einstein’s Explanation
Light consists of quanta called photons, each carrying energy $$E = h\nu$$ and momentum $$p = h/\lambda$$. A photon transfers all its energy to one electron:
$$h\nu = \phi + K_{\max}$$
where $$\phi = h\nu_0$$ is the work function of the metal.
3. Stopping Potential Method
In an experiment, the potential $$V_s$$ required to just stop the most energetic electrons is measured.
$$K_{\max} = eV_s$$
Combining,
$$eV_s = h\nu - h\nu_0 = h(\nu - \nu_0)$$
4. Typical JEE Graphs
- Plot of $$K_{\max}$$ vs $$\nu$$: Straight line, slope $$h$$, $$x$$-intercept $$\nu_0$$.
- Plot of $$V_s$$ vs $$\nu$$: Straight line with slope $$h/e$$.
- Photoelectric current vs intensity at constant $$\nu$$: straight line through origin.
- Photoelectric current vs frequency at constant intensity: zero below $$\nu_0$$, rises sharply above $$\nu_0$$.
5. Worked Example – Find Work Function
Question: A metal surface just stops photoelectrons when light of wavelength $$300\ \text{nm}$$ is incident. Calculate its work function in eV. ($$h = 6.63\times10^{-34}$$ J s, $$c = 3\times10^8$$ m/s, $$e = 1.6\times10^{-19}$$ C)
Solution:
Stopping means $$K_{\max}=0$$, so $$h\nu = \phi$$.
Frequency $$\nu = c/\lambda = 3\times10^8 / 3\times10^{-7} = 10^{15}\ \text{Hz}$$.
Energy $$\phi = h\nu = 6.63\times10^{-34} \times 10^{15} = 6.63\times10^{-19}\ \text{J}$$.
Convert to eV: $$\phi = \dfrac{6.63\times10^{-19}}{1.6\times10^{-19}} \approx 4.14\ \text{eV}$$.
Answer: 4.14 eV
6. Practice Resources
Tackle more questions on the photoelectric effect and build speed from the curated set of JEE Questions. Full-length mocks are available at JEE Mains Mock Test.
de Broglie Waves and Matter Diffraction
1. de Broglie Hypothesis
Any moving particle of momentum $$p$$ has an associated wavelength
$$\lambda = \dfrac{h}{p} = \dfrac{h}{mv}$$.
For an electron accelerated by potential $$V$$, $$p = \sqrt{2meV}$$, hence
$$\lambda = \dfrac{h}{\sqrt{2meV}}$$.
2. Davisson–Germer Experiment
- Electron beam accelerated by 54 V falls on nickel crystal.
- Strong diffraction peak observed at angle as predicted by Bragg’s law for $$\lambda \approx 0.167\ \text{nm}$$, matching de Broglie value.
- Confirmed wave nature of electrons.
3. Wave Packets and Group Velocity
Actual particle is represented by a wave packet. Group velocity $$v_g$$ equals particle velocity $$v$$ for non-relativistic speeds.
4. Electron Microscope
Resolution $$\propto \lambda$$ so shorter de Broglie wavelength of electrons gives higher resolving power than visible light. That is why electron microscopes can see atoms.
5. Worked Example – Electron Wavelength
Question: Calculate the de Broglie wavelength of electrons accelerated through 150 V.
Solution:
Use $$\lambda = \dfrac{h}{\sqrt{2meV}}$$.
$$h = 6.63\times10^{-34}$$ J s, $$m = 9.11\times10^{-31}$$ kg, $$e = 1.6\times10^{-19}$$ C, $$V = 150$$.
Denominator $$\sqrt{2meV} = \sqrt{2 \times 9.11\times10^{-31}\times1.6\times10^{-19}\times150}$$.
Compute inside: $$2meV= 2 \times 9.11 \times1.6 \times150 \times10^{-31-19} = 4368\times10^{-50} = 4.368\times10^{-47}$$.
Square root $$\approx 6.612\times10^{-24}$$ kg m/s.
Therefore $$\lambda = 6.63\times10^{-34} / 6.612\times10^{-24} \approx 1.00\times10^{-10}\ \text{m} = 0.10\ \text{nm}$$.
Answer: 0.10 nm
6. Relativistic Correction (when $$v$$ approaches $$c$$)
Use $$p = \gamma mv$$ with $$\gamma = 1/\sqrt{1 - v^2/c^2}$$. The general expression becomes $$\lambda = h/p$$ unchanged, but compute $$p$$ relativistically.
7. Practice Resources
For mixed modern-physics sets, solve papers in JEE Mains Previous Papers and JEE Advanced Previous Papers. Record formulae in your own sheet or download JEE Formula Sheets for last-minute checks.
Important Formulas and Results at a Glance
| Concept | Formula / Result | Units |
|---|---|---|
| Photon energy | $$E = h\nu = hc/\lambda$$ | J or eV |
| Photon momentum | $$p = h/\lambda = E/c$$ | kg m s-1 |
| Einstein’s photoelectric | $$h\nu = \phi + K_{\max}$$ | J or eV |
| Stopping potential | $$eV_s = K_{\max}$$ | J |
| de Broglie wavelength (general) | $$\lambda = h/p$$ | m |
| Electron accelerated by $$V$$ | $$\lambda = h/\sqrt{2meV}$$ | m |
| de Broglie for gas at $$T$$ | $$\lambda = h/\sqrt{3mkT}$$ (most probable) | m |
| Photon pressure on surface | $$P = \dfrac{I}{c}$$ (absorbing), $$P = \dfrac{2I}{c}$$ (perfectly reflecting) | N m-2 |
JEE Important Points, Common Mistakes and Quick Revision
Quick Revision Pointers
- Below threshold frequency no photoelectric effect, regardless of intensity. Do not confuse with current.
- Changing intensity changes number of emitted electrons, not their kinetic energy.
- Plot of $$V_s$$ vs $$\nu$$ gives slope $$h/e$$ – a popular data-based question.
- Photon momentum questions often ask for pressure or force on a surface; remember $$F = P \times A$$.
- For de Broglie wavelength questions, check if speed is relativistic. When $$eV \gt 100\ \text{keV}$$ for electrons, switch to relativistic formula.
- Electron diffraction through crystal relates to Bragg’s law $$2d\sin\theta = n\lambda$$ – keep $$\lambda$$ in metres.
- Units mix-up: use eV for energies but Joules in SI formulas. Convert properly with $$1\ \text{eV} = 1.6\times10^{-19}\ \text{J}$$.
Common Mistakes
- Plugging $$\nu_0$$ instead of $$\nu - \nu_0$$ into Einstein’s equation, giving wrong $$K_{\max}$$.
- Forgetting to square-root when rearranging $$\lambda = h/\sqrt{2meV}$$.
- Using $$E = mc^2$$ for photon energy; photons have zero rest mass.
- Assuming intensity affects threshold frequency; it does not.
- Ignoring work function units: $$\phi$$ should match $$h\nu$$ units.
Last-Minute 60-Second Drill
- Write down the four core formulas $$E = h\nu$$, $$p = h/\lambda$$, $$eV_s = h(\nu - \nu_0)$$ and $$\lambda = h/\sqrt{2meV}$$.
- Mental conversion: $$1240\ \text{eV nm}$$ is handy so $$\lambda(\text{nm}) = 1240/E(\text{eV})$$.
- Recall slope points: $$h$$ from $$K_{\max}\text{–}\nu$$ graph, $$h/e$$ from $$V_s\text{–}\nu$$ graph.
- Remember that doubling accelerating voltage reduces electron wavelength by $$1/\sqrt{2}$$.
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Dual Nature of Matter and Radiation JEE Notes: Conclusion
Dual nature of matter and radiation binds photon concepts and quantum mechanics into one scoring JEE chapter. Remember Einstein’s photoelectric equation for energy balance and de Broglie relation for wavelength of particles. Analyse graphs, keep units consistent, and practise a variety of numerical levels. A firm grasp here will secure easy marks with minimal calculation time in the exam.
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