Dual Nature of Matter and Radiation
Dual Nature of Matter and Radiation for JEE Main & Advanced
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:
- Threshold Frequency $\nu_0$: Minimum frequency below which no photoelectric emission occurs, regardless of intensity. Different for different metals.
- Instantaneous Emission: No measurable time lag between incidence of light and electron emission ($\sim 10^{-9}$ s).
- Stopping Potential $V_0$: Reverse potential at which photocurrent becomes zero. Depends only on frequency, not intensity.
- Saturation Current: Photocurrent reaches saturation when all emitted electrons reach anode. Proportional to intensity.
- Stopping potential varies linearly with frequency: $V_0 = (h/e)\nu - (\phi/e)$.
Laws of Photoelectric Emission:
| Law | Statement |
|---|---|
| First | Photocurrent $\propto$ intensity (for $\nu > \nu_0$) |
| Second | Max KE of photoelectrons depends on frequency, not intensity |
| Third | Photoelectric emission stops below $\nu_0$, regardless of intensity |
| Fourth | Emission 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.
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.
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.
Photoelectric effect was first observed by:
The stopping potential depends on:
Maximum KE of photoelectrons depends on:
Below threshold frequency:
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:
| Property | Formula |
|---|---|
| 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 $\lambda$: $$N = \frac{P}{E_{\text{photon}}} = \frac{P\lambda}{hc}$$
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.
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.
Energy of a photon of frequency $\nu$:
Momentum of photon:
Slope of $V_0$ vs $\nu$ graph:
The energy in eV of a photon of $\lambda = 248$ nm:
Photoelectric current is independent of:
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:
| Property | Photon | Matter 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.
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.
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).
The de Broglie wavelength of a particle:
de Broglie wavelength of electron at $100$ V:
The Heisenberg uncertainty principle relates:
Matter waves are significant at:
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:
| Application | Use |
|---|---|
| Electron microscope | Higher resolution than optical microscope (smaller $\lambda$ → finer detail) |
| Neutron diffraction | Studying crystal structure (neutrons have suitable $\lambda$ at thermal energies) |
| Scanning tunneling microscope | Atomic-scale imaging |
| Particle accelerators | Use 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.
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.
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.
The Davisson-Germer experiment established:
The de Broglie wavelength of electron in Davisson-Germer ($V = 54$ V):
The electron microscope has high resolution because:
Compton effect demonstrates:
Bragg's law:
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