Dual Nature of Radiation and Matter: Class 12 Physics NCERT Guide

The Photoelectric Effect: Light Behaving as a Particle

The photoelectric effect is the emission of electrons from a metal surface when light of a suitable frequency shines on it. Classical wave theory predicted that if light intensity were increased, electrons would gain more energy and be ejected, regardless of the light's frequency. However, experiments showed something completely different.

Key Observations that Baffled Classical Physics:

1. Threshold Frequency (ν₀): For a given metal, there exists a minimum frequency of incident light below which no photoelectrons are emitted, no matter how intense the light is. Wave theory had no explanation for this.
2. Instantaneous Emission: Electrons are ejected almost instantaneously (in less than 10⁻⁹ s) after the light hits the surface, even for very weak light, as long as the frequency is above the threshold. Wave theory predicted a time lag for the electron to absorb enough energy.
3. Kinetic Energy and Frequency: The maximum kinetic energy of the emitted electrons depends only on the frequency of the incident light, not its intensity. Higher frequency means higher kinetic energy.
4. Intensity and Photocurrent: Increasing the intensity of light (at a frequency above ν₀) only increases the number of photoelectrons emitted per second (the photoelectric current), not their individual kinetic energy.

These observations could only be explained by assuming light is not a continuous wave but a stream of discrete energy packets called photons.

Key Definitions and Equations

Photon

A discrete packet or quantum of electromagnetic radiation. The energy of a photon is given by E = hν, where 'h' is Planck's constant and 'ν' is the frequency of the radiation.

Work Function (φ₀ or W)

The minimum amount of energy required to remove an electron from the surface of a given metal. It is a property of the material. Measured in electron-volts (eV) or Joules.

Threshold Frequency (ν₀)

The minimum frequency of incident radiation required to cause photoemission from a particular metal surface. It is related to the work function by φ₀ = hν₀.

Stopping Potential (V₀)

The minimum negative (retarding) potential applied to the collector plate that just stops the photoelectric current, i.e., it stops even the most energetic photoelectrons. K_max = eV₀.

Einstein's Photoelectric Equation

Incident Photon Energy = Work Function + Max. Kinetic Energy of Photoelectron. Mathematically: hν = φ₀ + K_max or K_max = hν - φ₀.

de Broglie Wavelength (λ)

The wavelength associated with a moving particle. It is given by λ = h/p = h/mv, where 'h' is Planck's constant, 'p' is the momentum, 'm' is the mass, and 'v' is the velocity of the particle.

Worked Examples: Calculations in Action

• Example 1: Photoelectric Effect Calculation Light of wavelength 400 nm is incident on a metal surface with a work function of 2.1 eV. Calculate (a) the energy of the incident photon in eV, (b) the maximum kinetic energy of the photoelectrons, and (c) the stopping potential. (Given: h = 6.63 x 10⁻³⁴ Js, c = 3 x 10⁸ m/s, 1 eV = 1.6 x 10⁻¹⁹ J) Step 1: Calculate the energy of the incident photon. First, find the energy in Joules: E = hc/λ E = (6.63 x 10⁻³⁴ Js * 3 x 10⁸ m/s) / (400 x 10⁻⁹ m) = 4.97 x 10⁻¹⁹ J Now, convert this energy to electron-volts (eV): E (in eV) = (4.97 x 10⁻¹⁹ J) / (1.6 x 10⁻¹⁹ J/eV) ≈ 3.11 eV Step 2: Calculate the maximum kinetic energy (K_max). Use Einstein's photoelectric equation: K_max = E - φ₀ K_max = 3.11 eV - 2.1 eV = 1.01 eV Step 3: Calculate the stopping potential (V₀). The maximum kinetic energy is related to the stopping potential by K_max = eV₀. So, 1.01 eV = eV₀ This directly gives V₀ = 1.01 V. Final Answer: (a) 3.11 eV, (b) 1.01 eV, (c) 1.01 V.
• Example 2: de Broglie Wavelength Calculation Calculate the de Broglie wavelength of (a) an electron moving with a speed of 6 x 10⁶ m/s and (b) a cricket ball of mass 150 g moving at 30 m/s. (Given: mass of electron mₑ = 9.1 x 10⁻³¹ kg, h = 6.63 x 10⁻³⁴ Js) (a) For the electron: Step 1: Identify the formula and variables. The de Broglie wavelength is λ = h/mv. Here, m = 9.1 x 10⁻³¹ kg and v = 6 x 10⁶ m/s. Step 2: Calculate the momentum (p = mv). p = (9.1 x 10⁻³¹ kg) (6 x 10⁶ m/s) = 54.6 x 10⁻²⁵ kg m/s Step 3: Calculate the wavelength (λ). λ = (6.63 x 10⁻³⁴ Js) / (54.6 x 10⁻²⁵ kg m/s) ≈ 0.121 x 10⁻⁹ m = 0.121 nm This wavelength is in the X-ray range and is measurable. (b) For the cricket ball: Step 1: Identify the formula and variables. Use λ = h/mv. Here, m = 150 g = 0.15 kg and v = 30 m/s. Step 2: Calculate the momentum (p = mv). p = (0.15 kg) (30 m/s) = 4.5 kg m/s Step 3: Calculate the wavelength (λ). λ = (6.63 x 10⁻³⁴ Js) / (4.5 kg m/s) ≈ 1.47 x 10⁻³⁴ m This wavelength is incredibly small, far too small to be detected by any instrument. This is why macroscopic objects do not exhibit wave-like properties in our daily experience. Final Answer: (a) 0.121 nm, (b) 1.47 x 10⁻³⁴ m.

De Broglie's Hypothesis: The Wave Nature of Matter

After Einstein showed that waves (light) could act like particles (photons), the French physicist Louis de Broglie had a moment of profound insight. In 1924, he proposed that nature loves symmetry. If waves can behave like particles, then shouldn't particles, like electrons, protons, and even baseballs, exhibit wave-like properties? He postulated that every moving particle has a 'matter wave' associated with it. The wavelength of this wave, now called the de Broglie wavelength, is inversely proportional to the particle's momentum.

λ = h / p = h / mv

Here, 'λ' is the de Broglie wavelength, 'h' is Planck's constant, and 'p' is the momentum (mass 'm' times velocity 'v') of the particle. This revolutionary idea was initially just a hypothesis. However, in 1927, Clinton Davisson and Lester Germer experimentally confirmed the wave nature of electrons by observing that a beam of electrons scattered by a nickel crystal showed a diffraction pattern, a characteristic hallmark of waves. This confirmed de Broglie's hypothesis and opened the door to the development of quantum mechanics and technologies like the electron microscope.

Exam Traps and Important Points

Pay close attention to these common areas of confusion in your board exams:

• Intensity vs. Frequency: This is the most common trap. Remember:
• Intensity of light (number of photons) affects the photoelectric current (number of electrons emitted).
• Frequency of light (energy of each photon) affects the maximum kinetic energy of the emitted electrons and thus the stopping potential.
• Units Conversion: Be extremely careful with units. Energy can be in Joules (J) or electron-volts (eV). Ensure you convert everything to a consistent system (usually SI units) before calculating. Remember 1 eV = 1.6 x 10⁻¹⁹ J.
• Threshold Condition: Always check if the incident photon energy (hν) is greater than the work function (φ₀). If hν < φ₀, there will be no photoemission, and K_max is not just negative, it's non-existent. The kinetic energy cannot be negative.
• de Broglie Wavelength Formula: When calculating de Broglie wavelength for a particle accelerated by a potential V, you can use the derived formula λ = h / √(2mqV), where 'q' is the charge of the particle. This is a useful shortcut for electrons.

Practice Questions with Solutions

• Q: The work function of caesium is 2.14 eV. Find the threshold frequency for caesium. A: Step 1: State the relationship between work function (φ₀) and threshold frequency (ν₀). The formula is φ₀ = hν₀. Step 2: Convert the work function from eV to Joules for SI consistency. φ₀ = 2.14 eV * (1.6 x 10⁻¹⁹ J/eV) = 3.424 x 10⁻¹⁹ J. Step 3: Rearrange the formula to solve for ν₀: ν₀ = φ₀ / h. Step 4: Substitute the values and calculate. ν₀ = (3.424 x 10⁻¹⁹ J) / (6.63 x 10⁻³⁴ Js) = 5.16 x 10¹⁴ Hz. Final answer: The threshold frequency for caesium is 5.16 x 10¹⁴ Hz.
• Q: What is the de Broglie wavelength of an electron with kinetic energy of 120 eV? A: Step 1: Relate kinetic energy (K) to momentum (p). The formula is K = p²/2m, so p = √(2mK). Step 2: Substitute this into the de Broglie wavelength formula λ = h/p. This gives λ = h / √(2mK). Step 3: Convert the kinetic energy to Joules. K = 120 eV (1.6 x 10⁻¹⁹ J/eV) = 1.92 x 10⁻¹⁷ J. Step 4: Substitute all values into the combined formula. Use m = 9.1 x 10⁻³¹ kg for an electron. λ = (6.63 x 10⁻³⁴ Js) / √(2 9.1 x 10⁻³¹ kg * 1.92 x 10⁻¹⁷ J) = (6.63 x 10⁻³⁴) / √(34.944 x 10⁻⁴⁸) = (6.63 x 10⁻³⁴) / (5.91 x 10⁻²⁴) ≈ 1.12 x 10⁻¹⁰ m. Final answer: The de Broglie wavelength is 1.12 x 10⁻¹⁰ m or 0.112 nm.
• Q: In a photoelectric effect experiment, what is the effect on photoelectric current and stopping potential if the intensity of incident light is doubled while the frequency is kept constant? A: Step 1: Analyze the effect of intensity. The intensity of light is proportional to the number of photons incident per unit area per unit time. Step 2: Relate intensity to photoelectric current. Doubling the number of incident photons (above threshold frequency) will double the number of electrons emitted per second. Therefore, the photoelectric current will be doubled. Step 3: Analyze the effect of frequency on stopping potential. The stopping potential (V₀) is determined by the maximum kinetic energy of the photoelectrons (K_max = eV₀). K_max depends only on the frequency of the incident light (K_max = hν - φ₀). Step 4: Conclude the effect on stopping potential. Since the frequency of the light is kept constant, the maximum kinetic energy of the photoelectrons does not change. Therefore, the stopping potential will remain unchanged. Final answer: The photoelectric current will be doubled, and the stopping potential will remain unchanged.
• Q: An alpha particle and a proton are accelerated from rest by the same potential. Find the ratio of their de Broglie wavelengths. A: Step 1: Use the de Broglie wavelength formula for a particle accelerated by a potential V: λ = h / √(2mqV). Step 2: Write the expressions for the proton (p) and alpha particle (α). λ_p = h / √(2m_p q_p V) and λ_α = h / √(2m_α q_α V). Note that V is the same for both. Step 3: Establish the relationship between the masses and charges. An alpha particle is a helium nucleus (2 protons, 2 neutrons). So, m_α ≈ 4m_p and q_α = 2q_p. Step 4: Find the ratio λ_p / λ_α. λ_p / λ_α = [h / √(2m_p q_p V)] / [h / √(2m_α q_α V)] = √(2m_α q_α V) / √(2m_p q_p V) = √(m_α q_α / m_p q_p). Step 5: Substitute the relationships from Step 3. λ_p / λ_α = √((4m_p 2q_p) / (m_p q_p)) = √8 = 2√2. Final answer: The ratio of the de Broglie wavelength of the proton to the alpha particle is 2√2 : 1.

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