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Dual Nature of Matter and Radiation Formula For JEE 2026
The chapter of Dual Nature of Matter and Radiation is significant to JEE Mains, and it tends to provide 1-2 questions annually. It contains such major points as the photoelectric effect, photon properties, de Broglie wavelength, and DavissonGermer experiment. Majority of questions are concept and straightforward and hence, having a clear knowledge of formulas and simple concepts may assist you score with ease.
This chapter can be a dependable scoring zone with the practice and the ability to revise it quickly. A JEE Mains Physics Formula PDF will allow you to review relevant formulas in a single location and achieve a higher rate of speed and accuracy in the exam.
Introduction: What is the Dual Nature?
In everyday life, we think of light as a wave (it shows interference and diffraction) and particles like electrons as tiny balls. But experiments in the early 1900s revealed a stunning fact: light can also behave like a stream of particles, and particles like electrons can also behave like waves. This is called the dual nature of matter and radiation — everything in nature has both a wave character and a particle character.
This chapter focuses on two key discoveries: (1) light behaving as particles (photons), demonstrated by the photoelectric effect, and (2) matter behaving as waves, described by de Broglie's hypothesis.
Photon Energy and Momentum Formulas
In 1905, Albert Einstein proposed that light is not just a continuous wave — it also comes in tiny packets of energy called photons. Each photon is like a bullet of light, carrying a specific amount of energy that depends on the light's frequency.
Photon: A quantum (smallest packet) of electromagnetic radiation. It has energy and momentum but zero rest mass.
Properties of a Photon
- Energy: $$E = h\nu = \dfrac{hc}{\lambda}$$ where $$h$$ = Planck's constant = $$6.626 \times 10^{-34}$$ J s, $$\nu$$ = frequency, $$\lambda$$ = wavelength, $$c$$ = speed of light = $$3 \times 10^8$$ m/s
- Momentum: $$p = \dfrac{h}{\lambda} = \dfrac{h\nu}{c} = \dfrac{E}{c}$$
- Rest mass of a photon = 0 (photons only exist in motion at speed $$c$$)
- Effective (dynamic) mass: $$m = \dfrac{E}{c^2} = \dfrac{h\nu}{c^2} = \dfrac{h}{\lambda c}$$
A useful unit for photon energy is the electron volt (eV): 1 eV = $$1.6 \times 10^{-19}$$ J. Visible light photons have energies around 1.8–3.1 eV.
Quick Formula: Energy in eV from Wavelength
$$$E \;(\text{in eV}) = \frac{12400}{\lambda \;(\text{in \AA})} = \frac{1240}{\lambda \;(\text{in nm})}$$$
This shortcut avoids plugging in $$h$$ and $$c$$ every time.
Worked Example
Find the energy and momentum of a photon of wavelength 6200 Angstrom.
Step 1: Energy using the shortcut:
$$E = \dfrac{12400}{6200} = 2 \text{ eV} = 2 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-19}$$ J
Step 2: Momentum:
$$p = \dfrac{E}{c} = \dfrac{3.2 \times 10^{-19}}{3 \times 10^8} = 1.07 \times 10^{-27}$$ kg m/s
Tip: Memorise $$E = 12400/\lambda$$ (with $$\lambda$$ in Angstrom) or $$E = 1240/\lambda$$ (with $$\lambda$$ in nm). This saves significant time in JEE numerical problems.
Photoelectric Effect: Einstein's Equation
When light of sufficiently high frequency falls on a metal surface, electrons are ejected from the surface. This phenomenon is called the photoelectric effect. The ejected electrons are called photoelectrons. This was one of the first clear pieces of evidence that light behaves as particles.
Photoelectric Effect: The emission of electrons from a metal surface when light of suitable frequency shines on it.
Work Function ($$\phi$$ or $$W_0$$): The minimum energy needed to remove an electron from the surface of a metal. Different metals have different work functions. Typical values: Caesium $$\approx$$ 2.14 eV, Aluminium $$\approx$$ 4.28 eV.
Threshold Frequency ($$\nu_0$$): The minimum frequency of light below which no photoelectric emission occurs, no matter how intense the light. Related to work function: $$\phi = h\nu_0$$.
Threshold Wavelength ($$\lambda_0$$): The maximum wavelength of light above which no emission occurs: $$\lambda_0 = c/\nu_0 = hc/\phi$$.
Key Experimental Observations
Several observations about the photoelectric effect could not be explained by the classical wave theory of light. These puzzles led Einstein to his photon theory.
Observations of the Photoelectric Effect
- Threshold frequency exists: Below $$\nu_0$$, no electrons are emitted regardless of light intensity.
- Instantaneous emission: Electrons are emitted within $$\sim 10^{-9}$$ s of light hitting the surface (no time delay).
- KE depends on frequency, not intensity: Increasing frequency increases the maximum kinetic energy of emitted electrons. Increasing intensity does not.
- Number of electrons depends on intensity: Brighter light ejects more electrons (larger photocurrent), but each electron's energy stays the same.
Einstein's Photoelectric Equation
Einstein explained the photoelectric effect using the photon model: each photon gives all its energy $$h\nu$$ to a single electron. Part of this energy is used to overcome the work function $$\phi$$, and the rest becomes the electron's kinetic energy.
$$$h\nu = \phi + K_{\max}$$$
or equivalently:
$$$K_{\max} = h\nu - \phi = h(\nu - \nu_0)$$$
where:
- $$h\nu$$ = energy of the incident photon
- $$\phi = h\nu_0$$ = work function of the metal
- $$K_{\max} = \frac{1}{2}mv_{\max}^2$$ = maximum kinetic energy of emitted photoelectrons
Stopping Potential ($$V_0$$): The minimum negative potential applied to the collector that just stops the most energetic photoelectrons. At stopping potential, $$K_{\max} = eV_0$$.
Stopping Potential
From Einstein's equation:
$$$eV_0 = h\nu - \phi$$$
$$$V_0 = \frac{h\nu - \phi}{e} = \frac{h}{e}(\nu - \nu_0)$$$
- $$V_0$$ depends on frequency of light, not on intensity
- A graph of $$V_0$$ vs $$\nu$$ is a straight line with slope $$= h/e$$ and $$x$$-intercept $$= \nu_0$$
Worked Example
Light of frequency $$7.2 \times 10^{14}$$ Hz falls on a metal with work function 2.0 eV. Find (a) the maximum KE of photoelectrons, (b) the stopping potential.
Step 1: Energy of incident photon:
$$E = h\nu = 6.626 \times 10^{-34} \times 7.2 \times 10^{14} = 4.77 \times 10^{-19}$$ J $$= \dfrac{4.77 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.98$$ eV
Step 2: Maximum KE:
$$K_{\max} = E - \phi = 2.98 - 2.0 = 0.98$$ eV
Step 3: Stopping potential:
$$eV_0 = K_{\max} \Rightarrow V_0 = \dfrac{0.98 \text{ eV}}{e} = 0.98$$ V
Worked Example
When light of wavelength 400 nm falls on a metal, the stopping potential is 1.1 V. Find the work function and threshold wavelength.
Step 1: Energy of photon: $$E = \dfrac{1240}{400} = 3.1$$ eV
Step 2: Work function: $$\phi = E - eV_0 = 3.1 - 1.1 = 2.0$$ eV
Step 3: Threshold wavelength: $$\lambda_0 = \dfrac{1240}{\phi} = \dfrac{1240}{2.0} = 620$$ nm
Tip: If $$V_0$$ vs $$\nu$$ graph data is given for two frequencies, set up two equations $$eV_0 = h\nu - \phi$$ and solve simultaneously. The slope gives $$h/e$$ and the intercept gives $$\phi/e$$.
Effect of Intensity and Frequency
Summary: What Depends on What?
| Property | Depends on | Does NOT depend on |
|---|---|---|
| $$K_{\max}$$ of photoelectrons | Frequency | Intensity |
| Stopping potential $$V_0$$ | Frequency | Intensity |
| Photoelectric current | Intensity | Frequency (above $$\nu_0$$) |
| Emission (yes/no) | Frequency ($$\geq \nu_0$$) | Intensity |
Note: The photoelectric effect cannot be explained by the wave theory. Waves would predict: (1) any frequency should work if intensity is high enough, (2) there should be a time delay, (3) KE should depend on intensity. All three predictions are wrong. Only the photon model explains the observations.
de Broglie Wavelength Formula
In 1924, Louis de Broglie proposed a revolutionary idea: if light (a wave) can behave as particles, then particles (like electrons) should also behave as waves. He suggested that every moving particle has an associated wavelength, now called the de Broglie wavelength.
de Broglie Wavelength: The wavelength associated with a moving particle: $$\lambda = h/p$$, where $$p$$ is the momentum of the particle.
- General formula: $$\lambda = \dfrac{h}{p} = \dfrac{h}{mv}$$ where $$m$$ = mass, $$v$$ = velocity, $$p$$ = momentum
- In terms of kinetic energy: $$\lambda = \dfrac{h}{\sqrt{2mK}}$$ (since $$K = p^2/2m \Rightarrow p = \sqrt{2mK}$$)
- For electron accelerated through potential $$V$$: $$K = eV$$, so $$\lambda = \dfrac{h}{\sqrt{2meV}}$$
- Numerical shortcut for electrons: $$\lambda = \dfrac{12.27}{\sqrt{V}}$$ Angstrom where $$V$$ is the accelerating potential in volts
For everyday objects (like a ball), the de Broglie wavelength is incredibly tiny ($$\sim 10^{-34}$$ m) and undetectable. But for very light particles like electrons, the wavelength is comparable to atomic dimensions ($$\sim 10^{-10}$$ m), and wave behaviour becomes observable.
Worked Example
Find the de Broglie wavelength of an electron accelerated through 100 V.
Method 1 (Shortcut):
$$\lambda = \dfrac{12.27}{\sqrt{100}} = \dfrac{12.27}{10} = 1.227$$ Angstrom
Method 2 (From scratch):
$$\lambda = \dfrac{h}{\sqrt{2meV}} = \dfrac{6.626 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times 100}}$$
$$= \dfrac{6.626 \times 10^{-34}}{\sqrt{2.912 \times 10^{-47}}} = \dfrac{6.626 \times 10^{-34}}{5.396 \times 10^{-24}} = 1.228 \times 10^{-10}$$ m $$= 1.228$$ Angstrom
Worked Example
A proton and an electron have the same kinetic energy. Which has a larger de Broglie wavelength?
$$\lambda = \dfrac{h}{\sqrt{2mK}}$$
Since $$K$$ is the same for both, $$\lambda \propto \dfrac{1}{\sqrt{m}}$$.
The electron has a much smaller mass than the proton, so the electron has a larger wavelength.
$$\dfrac{\lambda_e}{\lambda_p} = \sqrt{\dfrac{m_p}{m_e}} = \sqrt{\dfrac{1.67 \times 10^{-27}}{9.1 \times 10^{-31}}} \approx \sqrt{1836} \approx 42.8$$
The electron's wavelength is about 43 times larger.
de Broglie Wavelength in Terms of Temperature
For a particle in thermal equilibrium at temperature $$T$$:
$$$\lambda = \frac{h}{\sqrt{3mkT}}$$$
where $$k = 1.38 \times 10^{-23}$$ J/K is Boltzmann's constant and $$m$$ is the mass of the particle.
(This uses the average thermal kinetic energy: $$K = \frac{3}{2}kT$$.)
Tip: Remember: $$\lambda = 12.27/\sqrt{V}$$ Angstrom for electrons. JEE frequently asks for the de Broglie wavelength of electrons accelerated through a given voltage. This shortcut saves time.
Davisson–Germer Experiment: Wave Nature of Electrons
The Davisson–Germer experiment (1927) provided the first direct experimental proof that electrons behave as waves, confirming de Broglie's hypothesis.
How the Experiment Works
Davisson and Germer accelerated electrons through a known potential and directed them onto a nickel crystal surface. They measured the intensity of scattered (reflected) electrons at various angles.
They found that at certain angles, the intensity of scattered electrons showed sharp peaks (maxima) — exactly like the diffraction pattern you would expect from waves scattering off a crystal lattice. This proved electrons have wave nature.
Davisson–Germer Key Result
The diffraction condition (Bragg's law for electrons):
$$$n\lambda = d\sin\phi$$$
where:
- $$d$$ = spacing between atomic planes in the crystal
- $$\phi$$ = angle of scattering
- $$n$$ = order of diffraction (1, 2, 3, ...)
- $$\lambda$$ = de Broglie wavelength of electrons
The measured $$\lambda$$ matched the predicted value $$\lambda = h/\sqrt{2meV}$$ perfectly.
Note: The Davisson–Germer experiment confirmed de Broglie's hypothesis experimentally. The wavelength measured from the diffraction pattern agreed with $$\lambda = h/p$$ calculated from the electron's momentum.
Electron Microscope
An electron microscope uses the wave nature of electrons instead of light waves to form images. Since electrons can have wavelengths much shorter than visible light (400–700 nm), electron microscopes can resolve much smaller features.
The resolving power of any microscope is limited by the wavelength used: shorter wavelength means finer detail. Electrons accelerated through ~100 V have $$\lambda \approx 1.2$$ Angstrom, which is thousands of times smaller than visible light wavelengths.
Electron Microscope vs Optical Microscope
| Feature | Optical Microscope | Electron Microscope |
|---|---|---|
| Uses | Light waves | Electron waves |
| Wavelength | ~4000–7000 Angstrom | ~1 Angstrom (at 150 V) |
| Resolving power | ~200 nm | ~0.1 nm |
| Lenses | Glass lenses | Magnetic lenses |
Wave–Particle Duality Summary
All entities in nature — light, electrons, protons, atoms — exhibit both wave and particle behaviour. Which behaviour dominates depends on the experiment:
When Does Each Nature Show?
- Particle nature dominates: Photoelectric effect, Compton effect, pair production — experiments involving energy/momentum exchange
- Wave nature dominates: Diffraction, interference — experiments involving propagation through slits or crystals
- A single experiment never shows both natures simultaneously (Bohr's complementarity principle)
Note: Wave–particle duality is universal, but it is only noticeable for microscopic particles (electrons, photons). For macroscopic objects, the wavelength is so tiny ($$\sim 10^{-34}$$ m) that wave effects are undetectable.
Important Photoelectric Effect Graphs for JEE
Graphs to Remember for JEE
- $$V_0$$ vs $$\nu$$: Straight line, slope $$= h/e$$, $$x$$-intercept $$= \nu_0$$, $$y$$-intercept $$= -\phi/e$$
- $$K_{\max}$$ vs $$\nu$$: Straight line, slope $$= h$$, $$x$$-intercept $$= \nu_0$$
- Photocurrent vs $$V$$: Starts at $$-V_0$$ (stopping potential), rises and saturates at saturation current
- Photocurrent vs intensity: Straight line through origin (for fixed $$\nu > \nu_0$$)
- $$\lambda$$ vs $$V$$: Hyperbolic decrease ($$\lambda \propto 1/\sqrt{V}$$) for de Broglie wavelength
Worked Example
The stopping potential for a metal is 1.85 V for light of wavelength 3000 Angstrom and 0.82 V for wavelength 4000 Angstrom. Find $$h$$ and the work function.
Step 1: Write Einstein's equation for both cases:
$$eV_1 = \dfrac{hc}{\lambda_1} - \phi$$ and $$eV_2 = \dfrac{hc}{\lambda_2} - \phi$$
Step 2: Subtract to eliminate $$\phi$$:
$$e(V_1 - V_2) = hc\left(\dfrac{1}{\lambda_1} - \dfrac{1}{\lambda_2}\right)$$
$$1.6 \times 10^{-19}(1.85 - 0.82) = h \times 3 \times 10^8 \left(\dfrac{1}{3 \times 10^{-7}} - \dfrac{1}{4 \times 10^{-7}}\right)$$
$$1.6 \times 10^{-19} \times 1.03 = h \times 3 \times 10^8 \times 8.33 \times 10^{5}$$
$$h = \dfrac{1.648 \times 10^{-19}}{2.5 \times 10^{14}} = 6.59 \times 10^{-34}$$ J s
Step 3: Find $$\phi$$ using either equation:
$$\phi = \dfrac{hc}{\lambda_1} - eV_1 = \dfrac{6.59 \times 10^{-34} \times 3 \times 10^8}{3 \times 10^{-7}} - 1.6 \times 10^{-19} \times 1.85$$
$$= 6.59 \times 10^{-19} - 2.96 \times 10^{-19} = 3.63 \times 10^{-19}$$ J $$= 2.27$$ eV
Tip: For JEE problems with two $$V_0$$–$$\lambda$$ pairs, always subtract the two equations to eliminate $$\phi$$ first. This is the fastest approach and avoids unnecessary calculations.
Dual Nature of Matter and Radiation Formulas For JEE 2026: Conclusion
Formula of JEE 2026 Dual Nature of Matter and Radiation is very significant in terms of revision and performance in exams. As it has been mentioned, photon energy, photoelectric effect, equation provided by Einstein, stopping potential, de Broglie wavelength and Davisson-Germer experiment are discussed in this chapter. Apart from the fact that the majority of the questions are straightforward and conceptual, a clear knowledge of formulas and general concepts can make you solve the questions in a fast and accurate manner. This chapter can be one of the easiest scoring sections in physics in JEE with a proper revision.
During the last stage of preparation, concentrate on memorising the significant formulas, basic graphs, and all the main distinctions between the nature of waves and the nature of particles. Notice also shortcuts such as energy in electron volt and de Broglie wavelength of electrons accelerated by a potential difference. The time savings during the exam is one of the benefits of rehearsing and cleverly revising Dual Nature of Matter and Radiation formula in JEE 2026 to enhance your overall score. Good basics and frequent revision can make you very good in this chapter.
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