JEE Main & Advanced

Dual Nature of Matter and Radiation

Dual Nature of Matter and Radiation for JEE Main & Advanced

1
Module 1

Photoelectric Effect

Photoelectric Effect Observations and LawsTopic 1

Photoelectric Effect: Emission of electrons (photoelectrons) from a metal surface when light of suitable frequency falls on it. Discovered by Heinrich Hertz (1887); systematically studied by Hallwachs and Lenard.

Key Experimental Observations:

  1. Threshold Frequency $\nu_0$: Minimum frequency below which no photoelectric emission occurs, regardless of intensity. Different for different metals.
  2. Instantaneous Emission: No measurable time lag between incidence of light and electron emission ($\sim 10^{-9}$ s).
  3. Stopping Potential $V_0$: Reverse potential at which photocurrent becomes zero. Depends only on frequency, not intensity.
  4. Saturation Current: Photocurrent reaches saturation when all emitted electrons reach anode. Proportional to intensity.
  5. Stopping potential varies linearly with frequency: $V_0 = (h/e)\nu - (\phi/e)$.

Laws of Photoelectric Emission:

LawStatement
FirstPhotocurrent $\propto$ intensity (for $\nu > \nu_0$)
SecondMax KE of photoelectrons depends on frequency, not intensity
ThirdPhotoelectric emission stops below $\nu_0$, regardless of intensity
FourthEmission is instantaneous

Failure of Classical Wave Theory:

  • Cannot explain threshold frequency (wave intensity should give energy regardless of $\nu$)
  • Cannot explain instantaneous emission
  • Cannot explain why KE is independent of intensity
  • Predicts time delay for weak intensity

Work Function $\phi$: Minimum energy required to liberate electron from metal surface. $\phi = h\nu_0$, where $\nu_0$ = threshold frequency. Typically a few eV: $\phi_{\text{Cs}} = 1.8$ eV, $\phi_{\text{Pt}} = 6.4$ eV.

Worked Examples
1

Threshold wavelength for cesium is $660$ nm. Find work function in eV.

Show solution

$$\phi = \frac{hc}{\lambda_0} = \frac{1240 \text{ eV·nm}}{660 \text{ nm}} \approx 1.88 \text{ eV}$$

Final Answer: $\phi \approx 1.88$ eV.

2

When light of $\lambda = 400$ nm illuminates a metal of $\phi = 2$ eV, find the stopping potential.

Show solution

$E_{\text{photon}} = hc/\lambda = 1240/400 = 3.1$ eV. $K_{\max} = E_{\text{photon}} - \phi = 3.1 - 2 = 1.1$ eV. $V_0 = K_{\max}/e = 1.1$ V.

Final Answer: $V_0 = 1.1$ V.

✎ Self-Check — 5 questions0 / 5
Q1.

Photoelectric effect was first observed by:

Q2.

The stopping potential depends on:

Q3.

Maximum KE of photoelectrons depends on:

Q4.

Below threshold frequency:

Q5.

Saturation photocurrent depends on:

Einstein's Photoelectric Equation and ApplicationsTopic 2

Einstein's Quantum Theory of Light (1905): Light consists of quanta (photons) of energy $E = h\nu$. In photoelectric emission, each photon delivers all its energy to one electron.

Einstein's Photoelectric Equation: $$h\nu = \phi + K_{\max}$$ $$K_{\max} = h\nu - \phi = h(\nu - \nu_0) = hc\left(\frac{1}{\lambda} - \frac{1}{\lambda_0}\right)$$

Equivalently: $K_{\max} = eV_0$ (in terms of stopping potential).

Photon Properties:

PropertyFormula
Energy$E = h\nu = hc/\lambda$
Momentum$p = E/c = h\nu/c = h/\lambda$
Rest mass$0$
Speed$c$ (in vacuum)
Effective mass$m = E/c^2 = h\nu/c^2 = h/(c\lambda)$

Useful constants:

  • $h = 6.626 \times 10^{-34}$ J·s = $4.136 \times 10^{-15}$ eV·s
  • $hc = 1240$ eV·nm $= 1240$ MeV·fm
  • $1$ eV $= 1.6 \times 10^{-19}$ J

Robert Millikan (1916) verified Einstein's equation; slope of $V_0$ vs $\nu$ graph = $h/e$. Used to measure $h$.

Number of Photons per second from source of power $P$ at wavelength $\lambda$: $$N = \frac{P}{E_{\text{photon}}} = \frac{P\lambda}{hc}$$

Worked Examples
1

Find the energy and momentum of a photon of $\lambda = 600$ nm.

Show solution

$$E = hc/\lambda = (6.626 \times 10^{-34})(3 \times 10^8)/(600 \times 10^{-9})$$ $$= 3.31 \times 10^{-19} \text{ J} \approx 2.07 \text{ eV}$$ $$p = h/\lambda = 6.626 \times 10^{-34}/(600 \times 10^{-9}) = 1.1 \times 10^{-27} \text{ kg·m/s}$$

Final Answer: $E \approx 2.07$ eV; $p \approx 1.1 \times 10^{-27}$ kg·m/s.

2

A $60$ W sodium lamp emits monochromatic light of $\lambda = 590$ nm. Find number of photons emitted per second (assume $100\%$ efficiency).

Show solution

$$E_{\text{photon}} = hc/\lambda = (6.626 \times 10^{-34})(3 \times 10^8)/(590 \times 10^{-9}) = 3.37 \times 10^{-19} \text{ J}$$ $$N = P/E = 60/(3.37 \times 10^{-19}) \approx 1.78 \times 10^{20} \text{ photons/s}$$

Final Answer: $N \approx 1.78 \times 10^{20}$ photons/s.

✎ Self-Check — 5 questions0 / 5
Q1.

Energy of a photon of frequency $\nu$:

Q2.

Momentum of photon:

Q3.

Slope of $V_0$ vs $\nu$ graph:

Q4.

The energy in eV of a photon of $\lambda = 248$ nm:

Q5.

Photoelectric current is independent of:

2
Module 2

Matter Waves

de Broglie Wavelength and Wave-Particle DualityTopic 1

de Broglie Hypothesis (1924): All moving matter has wave-like properties. The wavelength associated with a particle: $$\lambda = \frac{h}{p} = \frac{h}{mv}$$

This is the de Broglie wavelength, exhibiting wave-particle duality of matter (parallel to that of light).

de Broglie Wavelength for Particle of KE $K$: $$\lambda = \frac{h}{\sqrt{2mK}}$$

For charged particle accelerated through potential $V$: $K = qV$. $$\lambda = \frac{h}{\sqrt{2mqV}}$$

For electron accelerated through $V$ volts: $$\lambda_e = \frac{12.27}{\sqrt{V}} \text{ Å} = \frac{1.227}{\sqrt{V}} \text{ nm}$$

Comparison with Photon:

PropertyPhotonMatter particle
Speed$c$$v < c$
Rest mass$0$$m_0 \neq 0$
Energy$h\nu = hc/\lambda$$K + mc^2$ (rel.); $p^2/(2m)$ (non-rel.)
Momentum$h/\lambda$$mv$ or $h/\lambda$

Heisenberg Uncertainty Principle: $$\Delta x \cdot \Delta p \geq h/(4\pi) = \hbar/2$$

Position and momentum cannot be measured simultaneously with arbitrary precision. Similarly: $\Delta E \cdot \Delta t \geq \hbar/2$.

Useful approximations for de Broglie wavelength:

  • Electron at $100$ V: $\lambda \approx 1.227$ Å
  • Thermal neutron ($\sim 300$ K, $K \approx 0.025$ eV): $\lambda \approx 1.8$ Å
  • $1$ kg ball at $1$ m/s: $\lambda \approx 10^{-34}$ m (negligibly small)

Matter waves significant only at microscopic scales.

Worked Examples
1

Find de Broglie wavelength of electron accelerated through $100$ V.

Show solution

$$\lambda = \frac{12.27}{\sqrt{100}} = \frac{12.27}{10} = 1.227 \text{ Å}$$

Final Answer: $\lambda \approx 1.23$ Å.

2

Find de Broglie wavelength of a $0.05$ kg ball moving at $20$ m/s. Comment.

Show solution

$$\lambda = h/(mv) = (6.626 \times 10^{-34})/(0.05 \times 20) = 6.63 \times 10^{-34} \text{ m}$$

Extremely small — far less than any nuclear dimension. Wave nature negligible at macroscopic scales.

Final Answer: $\lambda \approx 6.63 \times 10^{-34}$ m (negligible).

✎ Self-Check — 5 questions0 / 5
Q1.

The de Broglie wavelength of a particle:

Q2.

de Broglie wavelength of electron at $100$ V:

Q3.

The Heisenberg uncertainty principle relates:

Q4.

Matter waves are significant at:

Q5.

de Broglie wavelength of particle of KE $K$ and mass $m$:

Davisson-Germer Experiment and ApplicationsTopic 2

Davisson-Germer Experiment (1927): Confirmed wave nature of electrons. They scattered low-energy electrons (54 eV) off nickel crystal and observed a pronounced peak at $\theta = 50°$ from incident direction, consistent with Bragg-like diffraction.

Experimental Setup:

  • Electron gun produces beam (accelerating voltage $V = 54$ V)
  • Beam hits nickel target
  • Scattered electrons detected at various angles
  • Sharp peak at $50°$ confirms diffraction

Result: Wavelength calculated from de Broglie ($\lambda = 12.27/\sqrt{54} = 1.67$ Å) matched wavelength from Bragg's diffraction formula → confirmed matter waves.

G. P. Thomson Experiment: Independent confirmation by passing electron beam through thin metal films; observed diffraction rings similar to X-ray diffraction. Won Nobel Prize 1937 (with Davisson).

Bragg's Law: $2d\sin\theta = n\lambda$, where $d$ = atomic plane spacing.

Applications of Matter Waves:

ApplicationUse
Electron microscopeHigher resolution than optical microscope (smaller $\lambda$ → finer detail)
Neutron diffractionStudying crystal structure (neutrons have suitable $\lambda$ at thermal energies)
Scanning tunneling microscopeAtomic-scale imaging
Particle acceleratorsUse wave properties in quantum mechanics calculations

Electron Microscope: Better resolution because $\lambda_{\text{electron}} \ll \lambda_{\text{light}}$. Can resolve atomic structures.

Compton Effect: Scattering of X-ray photons by electrons. Wavelength shift: $$\Delta\lambda = \frac{h}{m_e c}(1 - \cos\theta) = \lambda_C(1 - \cos\theta)$$ where $\lambda_C = h/(m_e c) = 0.0243$ Å is Compton wavelength of electron. Established particle nature of light.

Worked Examples
1

Davisson-Germer experiment uses electron accelerated through $54$ V. Find de Broglie wavelength.

Show solution

$$\lambda = 12.27/\sqrt{54} \approx 12.27/7.35 \approx 1.67 \text{ Å}$$

Final Answer: $\lambda \approx 1.67$ Å.

2

A photon of $\lambda = 0.1$ Å is scattered by an electron at $90°$. Find the wavelength shift.

Show solution

$$\Delta\lambda = \lambda_C(1 - \cos 90°) = 0.0243(1 - 0) = 0.0243 \text{ Å}$$

Final Answer: $\Delta\lambda \approx 0.024$ Å.

✎ Self-Check — 5 questions0 / 5
Q1.

The Davisson-Germer experiment established:

Q2.

The de Broglie wavelength of electron in Davisson-Germer ($V = 54$ V):

Q3.

The electron microscope has high resolution because:

Q4.

Compton effect demonstrates:

Q5.

Bragg's law:

Ready to test yourself?

Attempt the full timed mock tests — Main & Advanced level.

Start Mock Test 1 →