The late 19th century marked a transition in physics, moving from classical wave theories of electromagnetism to the discovery of discrete atomic constituents. While Maxwell's equations and Hertz's experiments confirmed the wave nature of light, investigations into electrical discharge through gases at low pressures revealed the particulate nature of electricity.
Historical Chronicle: Thomson & Millikan
Metals contain free electrons responsible for electrical conductivity, but these electrons cannot spontaneously escape the metal surface due to an attractive electrostatic pull from positive ions. To escape, an electron must acquire a minimum threshold energy.
The minimum energy required by an electron to escape from a metal surface is called the work function , denoted by $\phi_0$. It is measured in electron-volts (eV), where 1 eV = 1.602 &× 10 -19 J . The work function depends on the properties of the metal and its surface cleanliness.
| Emission Type | Energy Source | Physical Mechanism | Typical Application / Limits |
|---|---|---|---|
| Thermionic Emission | Thermal Energy | Heating the metal imparts sufficient thermal kinetic energy to free electrons to overcome the surface barrier. | Vacuum tubes, cathode ray filaments. |
| Field Emission | Strong Electric Field | Applying a very high electric field ($\approx 10^8$ V/m) pulls electrons out of the metal. | Spark plugs, field emission microscopes. |
| Photoelectric Emission | Electromagnetic Radiation | Illuminating the metal surface with light of suitable frequency causes immediate electron ejection. | Phototubes, solar cells, light sensors. |
The photoelectric effect is the emission of electrons (called photoelectrons) when light of a sufficiently high frequency falls on a photosensitive metal surface.
Using an evacuated glass/quartz tube containing a photosensitive plate (emitter C) and a collector plate (A), researchers analyzed how the photoelectric current varies with light intensity, potential difference, and frequency.
| Experimental Observation | Wave Theory Prediction | Quantum Reality |
|---|---|---|
| Instantaneous Emission | Significant time lag (hours) as energy distributes over the wavefront. | Instantaneous emission ($< 10^{-9}$ s) via one-to-one photon-electron collisions. |
| Existence of Threshold Frequency ($\nu_0$) | Light of any frequency should cause emission if intensity is high enough. | No emission occurs if the photon energy $h\nu$ is less than the work function $\phi_0$. |
| Kinetic Energy Independence of Intensity | Higher intensity (larger wave amplitude) should yield higher kinetic energy. | Maximum kinetic energy depends linearly on frequency, not on intensity. |
In 1905, Albert Einstein resolved the contradictions of wave theory by proposing a particle-like description of light, where electromagnetic radiation is quantized into packets of energy.
Historical Chronicle: Albert Einstein
When a photon of energy $h\nu$ strikes an electron in a metal, the energy is absorbed. Part of this energy is used to overcome the work function $\phi_0$, and the remainder becomes the maximum kinetic energy of the emitted electron:
$$K_{\text{max}} = h\nu - \phi_0$$Using the stopping potential relation $K_{\text{max}} = e V_0$, the equation can be rewritten as:
$$V_0 = \left(\frac{h}{e}\right)\nu - \frac{\phi_0}{e}$$This predicts a straight-line graph of $V_0$ versus $\nu$ with a slope of $h/e$, which is universal for all metals.
Mathematical Problem Statement: A monochromatic laser of frequency 6.0 &× 10 14 Hz produces a power of 2.0 &× 10 -3 W . Find (a) the energy of a photon in the beam, and (b) the average number of photons emitted per second.
Step-by-Step Algorithmic Solution Trace:
- Calculate photon energy using $E = h\nu$:$$E = (6.63 \times 10^{-34} \text{ J s}) \times (6.0 \times 10^{14} \text{ Hz}) = 3.98 \times 10^{-19} \text{ J}$$
- Determine the photon emission rate $N$ from power $P = N E$:$$N = \frac{P}{E} = \frac{2.0 \times 10^{-3} \text{ W}}{3.98 \times 10^{-19} \text{ J}} \approx 5.0 \times 10^{15} \text{ photons/s}$$
Final Evaluation Output: Photon energy is $3.98 \times 10^{-19}$ J and emission rate is $5.0 \times 10^{15}$ photons/s .
Mathematical Problem Statement: The work function of caesium is 2.14 eV . Find (a) its threshold frequency, and (b) the wavelength of incident light if the photocurrent is brought to zero by a stopping potential of 0.60 V .
Step-by-Step Algorithmic Solution Trace:
- Convert work function to Joules: $\phi_0 = 2.14 \times 1.6 \times 10^{-19}$ J = $3.424 \times 10^{-19}$ J. Calculate threshold frequency $\nu_0 = \phi_0 / h$:$$\nu_0 = \frac{3.424 \times 10^{-19} \text{ J}}{6.63 \times 10^{-34} \text{ J s}} \approx 5.16 \times 10^{14} \text{ Hz}$$
- Use Einstein's equation $e V_0 = \frac{hc}{\lambda} - \phi_0$ to isolate wavelength $\lambda$:$$\lambda = \frac{hc}{e V_0 + \phi_0} = \frac{(6.63 \times 10^{-34}) \times (3 \times 10^8)}{0.60 \text{ eV} + 2.14 \text{ eV}} = \frac{1.989 \times 10^{-25} \text{ J m}}{2.74 \times 1.6 \times 10^{-19} \text{ J}} \approx 454 \times 10^{-9} \text{ m}$$
Final Evaluation Output: Threshold frequency is $5.16 \times 10^{14}$ Hz and incident wavelength is $454$ nm .
The photoelectric effect and Compton scattering provide definitive proof that light behaves as a stream of localized packets of energy and momentum called photons .
If electromagnetic waves exhibit particle-like properties, then material particles should, by symmetry, exhibit wave-like properties under appropriate conditions.
Historical Chronicle: Louis de Broglie
The wavelength $\lambda$ associated with a particle of mass $m$ moving with velocity $v$ (momentum $p$) is given by the de Broglie relation :
$$\lambda = \frac{h}{p} = \frac{h}{mv}$$For macroscopic objects, the mass is large, making the de Broglie wavelength extremely small and undetectable. For subatomic particles like electrons, the wavelength is comparable to atomic spacings and is highly significant.
Mathematical Problem Statement: Calculate the de Broglie wavelength associated with (a) an electron moving at 5.4 &× 10 6 m/s , and (b) a ball of mass 150 g traveling at 30.0 m/s .
Step-by-Step Algorithmic Solution Trace:
- For the electron ($m = 9.11 \times 10^{-31}$ kg):$$\lambda_e = \frac{h}{m_e v_e} = \frac{6.63 \times 10^{-34} \text{ J s}}{(9.11 \times 10^{-31} \text{ kg}) \times (5.4 \times 10^6 \text{ m/s})} \approx 1.35 \times 10^{-10} \text{ m} = 0.135 \text{ nm}$$
- For the ball ($m = 0.150$ kg):$$\lambda_b = \frac{h}{m_b v_b} = \frac{6.63 \times 10^{-34} \text{ J s}}{(0.150 \text{ kg}) \times (30.0 \text{ m/s})} \approx 1.47 \times 10^{-34} \text{ m}$$
Final Evaluation Output: The electron's wavelength is $0.135$ nm (measurable), while the ball's wavelength is $1.47 \times 10^{-34}$ m (undetectably small).
| Physical Quantity | Symbol | Dimensions | SI Unit | Key Formula / Remarks |
|---|---|---|---|---|
| Planck's Constant | $h$ | $[ML^2T^{-1}]$ | J s | $E = h\nu$ |
| Stopping Potential | $V_0$ | $[ML^2T^{-3}A^{-1}]$ | V | $e V_0 = K_{\text{max}}$ |
| Work Function | $\phi_0$ | $[ML^2T^{-2}]$ | J or eV | $\phi_0 = h\nu_0$ |
| Threshold Frequency | $\nu_0$ | $[T^{-1}]$ | Hz | $\nu_0 = \phi_0 / h$ |
| de Broglie Wavelength | $\lambda$ | $[L]$ | m | $\lambda = h / p$ |