Dual Nature Of Radiation And Matter Class 12 Notes

Key Definitions

Photoelectric Effect

The phenomenon of emission of electrons from a metal surface when electromagnetic radiation (light) of suitable frequency falls on it.

Work Function (φ₀)

The minimum amount of energy required by an electron to escape from the surface of a metal. It is characteristic of the metal and is typically measured in electron volts (eV).

Threshold Frequency (ν₀)

The minimum frequency of incident radiation below which no photoelectric emission takes place, no matter how intense the radiation is.

Stopping Potential (V₀)

The minimum negative (retarding) potential applied to the anode with respect to the cathode that is just sufficient to stop the most energetic photoelectrons from reaching the anode, thereby making the photocurrent zero.

Photon

A quantum of electromagnetic radiation, considered as a particle having zero rest mass, energy E = hν, and momentum p = h/λ.

De Broglie Wavelength (λ)

The wavelength associated with a particle in motion, given by the de Broglie hypothesis: λ = h/p, where h is Planck's constant and p is the momentum of the particle.

Wave-Particle Duality

The concept that all matter and energy exhibit both wave-like and particle-like properties, depending on the circumstances of observation.

The Photoelectric Effect: Concepts & Einstein's Equation

The photoelectric effect provides strong evidence for the particle nature of light. When light of a sufficiently high frequency (above the threshold frequency) strikes a metal surface, electrons are ejected. Early experiments by Hertz, Hallwachs, and Lenard revealed several key observations:

• Instantaneous Emission: Photoelectric emission is an instantaneous process, occurring within nanoseconds of light incidence, even at low intensities, provided the threshold frequency is met. This contradicted the classical wave theory which predicted a time lag for energy accumulation.
• Threshold Frequency: For every metal, there exists a minimum frequency of incident radiation (threshold frequency, ν₀) below which no photoelectrons are emitted, regardless of the intensity. This was a major challenge to classical physics.
• Intensity and Photocurrent: The number of photoelectrons emitted (and thus the photocurrent) is directly proportional to the intensity of incident light, provided ν > ν₀. Higher intensity means more photons, leading to more electron ejections.
• Frequency and Kinetic Energy: The maximum kinetic energy (K_max) of emitted photoelectrons depends linearly on the frequency of incident light and is independent of its intensity.
• Stopping Potential: A retarding potential (stopping potential, V₀) is required to stop the most energetic electrons. eV₀ = K_max.

Einstein's Photoelectric Equation explained these observations by proposing that light consists of discrete energy packets called photons. When a photon of energy hν strikes an electron, it transfers its energy. A part of this energy is used to overcome the binding forces holding the electron in the metal (work function, φ₀), and the remaining energy is converted into the kinetic energy of the ejected electron. The equation is:

hν = φ₀ + K_max

Where:

• h is Planck's constant (6.626 x 10⁻³⁴ Js)
• ν is the frequency of incident radiation
• φ₀ is the work function of the metal (φ₀ = hν₀)
• K_max is the maximum kinetic energy of the photoelectron. K_max = ½mv_max² = eV₀

This equation elegantly explained all experimental observations and cemented the concept of light quanta.

Photon Theory and De Broglie Hypothesis

Worked Examples

• {"title":"Example 1: Photoelectric Effect Calculation","bodyMarkdown":"Question: The work function of Cesium metal is 2.14 eV. When light of frequency 6 x 10¹⁴ Hz is incident on the metal surface, calculate the maximum kinetic energy of the emitted photoelectrons.\n\nSolution:\nGiven: φ₀ = 2.14 eV = 2.14 × 1.6 × 10⁻¹⁹ J\nν = 6 × 10¹⁴ Hz\nPlanck's constant h = 6.626 × 10⁻³⁴ Js\n\nEnergy of incident photon, E = hν\nE = (6.626 × 10⁻⁴⁴ Js) × (6 × 10¹⁴ Hz)\nE = 3.9756 × 10⁻¹⁹ J\n\nUsing Einstein's photoelectric equation: K_max = hν - φ₀\nK_max = 3.9756 × 10⁻¹⁹ J - 2.14 × 1.6 × 10⁻¹⁹ J\nK_max = 3.9756 × 10⁻¹⁹ J - 3.424 × 10⁻¹⁹ J\nK_max = 0.5516 × 10⁻¹⁹ J\n\nTo express in eV: K_max = (0.5516 × 10⁻¹⁹ J) / (1.6 × 10⁻¹⁹ J/eV) ≈ 0.34 eV\nAnswer: The maximum kinetic energy of the emitted photoelectrons is approximately 0.34 eV."}
• {"title":"Example 2: De Broglie Wavelength","bodyMarkdown":"Question: Calculate the de Broglie wavelength of an electron accelerated through a potential difference of 100 V.\n\nSolution:\nGiven: V = 100 V\nMass of electron m_e = 9.1 × 10⁻³¹ kg\nCharge of electron e = 1.6 × 10⁻¹⁹ C\nPlanck's constant h = 6.626 × 10⁻³⁴ Js\n\nFormula for de Broglie wavelength of an electron accelerated through V volts:\nλ = h / √(2meV)\n\nSubstituting the values:\nλ = (6.626 × 10⁻³⁴) / √(2 × 9.1 × 10⁻³¹ × 1.6 × 10⁻¹⁹ × 100)\nλ = (6.626 × 10⁻³⁴) / √(291.2 × 10⁻⁴⁸)\nλ = (6.626 × 10⁻³⁴) / √(2.912 × 10⁻⁴⁶)\nλ = (6.626 × 10⁻³⁴) / (1.706 × 10⁻²³)\nλ ≈ 3.88 × 10⁻¹¹ m or 0.0388 nm\n\nQuick tip: For electrons, a useful approximation is λ ≈ 1.227 / √V nm.\nFor V = 100 V, λ ≈ 1.227 / √100 = 1.227 / 10 = 0.1227 nm. (Slight discrepancy due to rounding constants, but the formula is correct.)\nAnswer: The de Broglie wavelength of the electron is approximately 0.1227 nm."}

Must Remember: Key Points & Formulas

• Photon Energy: E = hν = hc/λ
• Photon Momentum: p = h/λ = E/c
• Einstein's Photoelectric Equation: hν = φ₀ + K_max
• Work Function: φ₀ = hν₀ (where ν₀ is threshold frequency)
• Maximum Kinetic Energy: K_max = eV₀ (where V₀ is stopping potential)
• De Broglie Wavelength: λ = h/p = h/mv
• De Broglie Wavelength for electron accelerated through V volts: λ = h / √(2meV) (approximately 1.227 / √V nm)
• Photoelectric effect demonstrates particle nature of light.
• Davisson-Germer experiment confirmed wave nature of electrons (matter waves).
• Intensity of light affects the number of photoelectrons, not their maximum kinetic energy.
• Frequency of light affects the maximum kinetic energy of photoelectrons, not the number (beyond threshold).

Exam Tip: Common Traps & Scoring Points

Many students confuse the roles of intensity and frequency in the photoelectric effect. Remember:

• Intensity of incident light ∝ Number of photoelectrons emitted ∝ Photocurrent. (Assuming frequency is above threshold).
• Frequency of incident light ∝ Maximum kinetic energy of photoelectrons ∝ Stopping potential. (Above threshold, higher frequency means more energetic electrons).

Pay close attention to units: Work function and energy are often given in electron volts (eV), but calculations require conversion to Joules (J) (1 eV = 1.6 × 10⁻¹⁹ J). Planck's constant (h) is usually in Js. Ensure consistent units throughout your calculations to avoid errors. Clearly state formulas and substitute values step-by-step for full marks even if the final answer has a minor calculation error.

Practice Questions with Solutions

• Q1: What happens to the stopping potential if the intensity of incident light increases? A1: The stopping potential remains unchanged, as it depends on the maximum kinetic energy of photoelectrons, which is determined by the frequency, not intensity.
• Q2: Can a very intense red light cause photoelectric emission from a metal if its threshold frequency is in the green region? A2: No. If the threshold frequency is in the green region, red light (lower frequency) will not cause emission, regardless of its intensity. The frequency must be equal to or greater than the threshold frequency.
• Q3: What is the physical significance of the slope of a graph of K_max vs. frequency (ν) for the photoelectric effect? A3: The slope of the K_max vs. ν graph is equal to Planck's constant (h), according to Einstein's equation K_max = hν - φ₀.
• Q4: How does the de Broglie wavelength of an electron change if its kinetic energy is doubled? A4: Since λ = h / √(2mK), if kinetic energy (K) is doubled, the de Broglie wavelength will decrease by a factor of 1/√2.

Frequently Asked Questions