Chapter 11: Dual Nature of Radiation and Matter Case Study Questions | physics Class 12 CBSE

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Chapter 11: Dual Nature of Radiation and Matter Case Study Questions | physics Class 12 CBSE | ByteNotes.in

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A physics teacher sets up a photoelectric experiment in the laboratory. A monochromatic light source of frequency 8.0 × 10¹⁴ Hz illuminates a caesium metal plate (work function 2.14 eV). The collector plate is connected to a variable voltage supply. When the collector is at a positive potential, a steady photocurrent is observed. The teacher gradually increases the retarding potential until the current drops to zero. She then repeats the experiment with a light source of the same frequency but three times the original intensity. Students are asked to predict and explain the outcomes. The experiment is designed to help students distinguish between what changes and what remains constant when only intensity is varied, and to understand the quantum nature of the photoelectric process.

Question 1: What is the maximum kinetic energy of the photoelectrons emitted from the caesium plate when light of frequency 8.0 × 10¹⁴ Hz is incident on it? (h = 6.63 × 10⁻³⁴ J s, 1 eV = 1.6 × 10⁻¹⁹ J)

Photon energy hν = 6.63 × 10⁻³⁴ × 8.0 × 10¹⁴ = 5.304 × 10⁻¹⁹ J = 3.315 eV. Work function φ₀ = 2.14 eV.
Kmax = hν − φ₀ = 3.315 − 2.14 = 1.175 eV ≈ 1.18 eV. This is the maximum kinetic energy of the emitted photoelectrons.

Question 2: When the intensity of the incident light is tripled (frequency kept constant), what happens to the stopping potential and the saturation current? Justify your answer.

The stopping potential remains unchanged because it depends only on the frequency of incident radiation and the work function (V₀ = (hν − φ₀)/e), neither of which changes when only intensity is varied.
The saturation current triples because higher intensity means three times as many photons per second, causing three times as many photoelectrons to be emitted per second, resulting in three times the saturation current.

Question 3: Explain why classical wave theory fails to account for the observations in this experiment, specifically addressing the existence of a definite stopping potential that is independent of intensity.

Wave theory treats light as a continuous electromagnetic wave whose energy is distributed uniformly over the wavefront. It predicts that greater intensity delivers more energy per unit time to each electron, so Kmax should increase with intensity. But the experiment shows Kmax (and hence stopping potential) is independent of intensity — a direct contradiction.
According to wave theory, since energy is absorbed continuously, any intense radiation should supply enough energy to electrons regardless of frequency, predicting no threshold frequency and no fixed stopping potential. Experimentally, the stopping potential is uniquely determined by frequency, not intensity, which the wave model cannot explain.
Einstein's photon model resolves this: each electron absorbs exactly one photon of energy hν. The maximum kinetic energy is Kmax = hν − φ₀, fixed by frequency alone. Intensity only determines how many electrons are emitted per second (photocurrent), not the energy of individual electrons, perfectly explaining the constant stopping potential.

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During a science exhibition, a student demonstrates the photoelectric effect using different metals. She uses a single UV light source emitting radiation of wavelength 200 nm. She illuminates plates made of zinc (work function 4.3 eV), sodium (work function 2.3 eV), and caesium (work function 2.14 eV) one at a time. A sensitive microammeter connected in the circuit registers photocurrent only for certain metals. The student also uses green light (wavelength 550 nm) and red light (wavelength 700 nm) to check sensitivity. She notes which metals respond to which light. Her findings demonstrate that photoelectric emission depends critically on the relationship between the incident photon energy and the metal's work function, and that different metals require different minimum frequencies of light.

Question 1: Calculate the energy of a photon of UV light with wavelength 200 nm. (h = 6.63 × 10⁻³⁴ J s, c = 3 × 10⁸ m/s, 1 eV = 1.6 × 10⁻¹⁹ J)

Energy E = hc/λ = (6.63 × 10⁻³⁴ × 3 × 10⁸) / (200 × 10⁻⁹) = 19.89 × 10⁻²⁶ / 200 × 10⁻⁹ = 9.945 × 10⁻¹⁹ J.
Converting to eV: E = 9.945 × 10⁻¹⁹ / 1.6 × 10⁻¹⁹ ≈ 6.22 eV. This photon energy exceeds the work function of all three metals listed.

Question 2: Which metals will show photoelectric emission under green light (wavelength 550 nm, photon energy ≈ 2.26 eV)? Justify using the concept of threshold frequency.

For photoelectric emission, photon energy must exceed the work function. Green light photon energy ≈ 2.26 eV. Zinc (4.3 eV) and sodium (2.3 eV) have work functions greater than or equal to 2.26 eV, so no emission occurs for zinc; sodium is borderline and effectively does not emit.
Only caesium (φ₀ = 2.14 eV) has a work function less than 2.26 eV, so green light photons can liberate electrons from caesium. The threshold frequency for caesium is lower than the frequency of green light, satisfying ν > ν₀.

Question 3: For zinc illuminated by UV light (photon energy 6.22 eV), calculate the maximum kinetic energy of the emitted photoelectrons and the corresponding stopping potential. Explain why increasing the UV intensity will not change the stopping potential.

Maximum kinetic energy: Kmax = hν − φ₀ = 6.22 eV − 4.3 eV = 1.92 eV = 1.92 × 1.6 × 10⁻¹⁹ = 3.072 × 10⁻¹⁹ J.
Stopping potential V₀ = Kmax/e = 1.92 eV / e = 1.92 V. This is the retarding potential at which the most energetic photoelectrons are just stopped and the photocurrent becomes zero.
Increasing UV intensity increases the number of photons per second but does not change the energy of each photon (hν is fixed by frequency). Each electron still absorbs one photon and gains the same Kmax = 1.92 eV. Hence stopping potential remains 1.92 V, confirming Einstein's photon model.

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Case Set 3

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A CBSE student reads about Millikan's experiments conducted between 1906 and 1916. Millikan meticulously measured the stopping potential for sodium metal at several frequencies of incident radiation. He plotted stopping potential on the y-axis against frequency on the x-axis and obtained a precise straight line. From the slope of this line, he calculated a value of Planck's constant that closely matched the value determined from blackbody radiation studies. Initially, Millikan set out to disprove Einstein's photoelectric equation. Instead, his precision measurements confirmed it beyond doubt. The slope of Millikan's V₀ versus frequency graph was found to be 4.14 × 10⁻¹⁵ V s, and the x-intercept (threshold frequency) for sodium was approximately 5.5 × 10¹⁴ Hz. These results became a cornerstone of quantum physics.

Question 1: What does the x-intercept of the stopping potential versus frequency graph represent, and what is the work function of sodium using the data given? (e = 1.6 × 10⁻¹⁹ C)

The x-intercept of the V₀ versus ν graph represents the threshold frequency ν₀ of the metal, the minimum frequency below which no photoelectric emission occurs and V₀ = 0.
Work function φ₀ = hν₀. Using slope = h/e = 4.14 × 10⁻¹⁵ V s, h = 4.14 × 10⁻¹⁵ × 1.6 × 10⁻¹⁹ = 6.624 × 10⁻³⁴ J s. So φ₀ = 6.624 × 10⁻³⁴ × 5.5 × 10¹⁴ ≈ 3.64 × 10⁻¹⁹ J ≈ 2.27 eV.

Question 2: Why is the slope of the V₀ versus frequency graph the same for all metals, while the x-intercept differs from metal to metal?

The slope of the V₀ versus ν graph is h/e, derived from Einstein's equation V₀ = (h/e)ν − φ₀/e. Since h (Planck's constant) and e (electron charge) are universal physical constants, the slope h/e is the same for all metals regardless of their chemical nature.
The x-intercept gives the threshold frequency ν₀ = φ₀/h, which depends on the work function φ₀. Since φ₀ varies from metal to metal (it is a material property depending on the nature and surface of the metal), the x-intercept differs for each metal.

Question 3: Millikan aimed to disprove Einstein's equation but instead confirmed it. Explain what Einstein's equation predicts about the V₀ versus frequency relationship, and discuss the significance of Millikan's confirmation for the acceptance of the photon model.

Einstein's equation eV₀ = hν − φ₀ predicts a strictly linear relationship between stopping potential V₀ and frequency ν, with slope h/e (universal constant) and y-intercept −φ₀/e (metal-specific). Below ν₀ = φ₀/h, no emission occurs (V₀ = 0 or undefined).
Millikan's precise measurements of the V₀ versus ν straight line for sodium, yielding a slope that gave h = 6.624 × 10⁻³⁴ J s (consistent with Planck's constant from blackbody radiation), confirmed every prediction of Einstein's equation with great accuracy over a wide frequency range.
This confirmation was pivotal: it provided the first independent experimental determination of Planck's constant via photoelectric measurements, validated the revolutionary concept of light quantisation (photons), and led to the acceptance of Einstein's photon model of electromagnetic radiation. Einstein received the Nobel Prize in 1921 and Millikan in 1923 for these connected contributions.

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In 1924, Louis de Broglie proposed that matter, like radiation, has a dual nature. He suggested that a particle of momentum p has an associated wavelength λ = h/p. A physics professor demonstrates this concept by comparing the de Broglie wavelengths of different objects: a cricket ball of mass 0.15 kg moving at 30 m/s, a proton moving at 10⁶ m/s, and an electron accelerated through a potential difference of 100 V. The professor emphasises that the wave nature of matter is only observable for subatomic particles because macroscopic objects have de Broglie wavelengths far smaller than any measurable scale. This concept underpins all of modern quantum mechanics and was confirmed experimentally by electron diffraction. De Broglie was awarded the Nobel Prize in 1929 for this discovery.

Question 1: Calculate the de Broglie wavelength of the cricket ball (mass 0.15 kg, speed 30 m/s). Why is this wavelength not observable? (h = 6.63 × 10⁻³⁴ J s)

Momentum p = mv = 0.15 × 30 = 4.5 kg m/s. De Broglie wavelength λ = h/p = 6.63 × 10⁻³⁴ / 4.5 ≈ 1.47 × 10⁻³⁴ m.
This wavelength is approximately 10⁻¹⁹ times the size of a proton, far beyond the resolution of any instrument. No physical slit or crystal can diffract waves of such short wavelength, making wave behaviour of the cricket ball completely unobservable.

Question 2: An electron is accelerated through a potential difference of 100 V. Calculate its de Broglie wavelength. (m = 9.11 × 10⁻³¹ kg, e = 1.6 × 10⁻¹⁹ C, h = 6.63 × 10⁻³⁴ J s)

Kinetic energy K = eV = 1.6 × 10⁻¹⁹ × 100 = 1.6 × 10⁻¹⁷ J. Momentum p = √(2mK) = √(2 × 9.11 × 10⁻³¹ × 1.6 × 10⁻¹⁷) = √(2.915 × 10⁻⁴⁷) ≈ 5.40 × 10⁻²⁴ kg m/s.
De Broglie wavelength λ = h/p = 6.63 × 10⁻³⁴ / 5.40 × 10⁻²⁴ ≈ 1.23 × 10⁻¹⁰ m = 0.123 nm. This is comparable to X-ray wavelengths and atomic spacings in crystals, making electron diffraction experimentally feasible.

Question 3: De Broglie's hypothesis was based on a symmetry argument between radiation and matter. Explain this argument and discuss how the de Broglie relation λ = h/p inherently embodies the dual nature of matter.

De Broglie argued that nature is fundamentally symmetrical. Radiation (energy) was shown to have dual wave-particle character: wave in interference/diffraction, particle (photon) in photoelectric and Compton effects. By symmetry, matter (particles) must also have dual character: particle in mechanical collisions, wave in quantum phenomena.
The relation λ = h/p is the mathematical expression of this duality. On the left side, λ (wavelength) is a purely wave attribute — it describes the spatial periodicity of a wave. On the right side, p (momentum) is a purely particle attribute — it is the product of mass and velocity of a localised body.
Planck's constant h is the bridge connecting these two worlds. Its extremely small value (6.626 × 10⁻³⁴ J s) means that wave behaviour is only significant when p is extremely small, i.e., for subatomic particles. This relation laid the foundation for Schrödinger's wave mechanics and all of modern quantum theory.

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A research team is designing a highly sensitive photomultiplier tube for a telescope to detect faint starlight. They need to select a photosensitive material with the lowest possible work function so that even low-energy visible photons can trigger electron emission. They consider four materials: platinum (φ₀ = 5.65 eV), zinc (φ₀ = 4.3 eV), sodium (φ₀ = 2.3 eV), and caesium (φ₀ = 2.14 eV). The telescope will primarily receive visible light in the frequency range 4 × 10¹⁴ Hz to 7 × 10¹⁴ Hz. The team also needs to ensure that the chosen material produces photoelectrons with sufficient kinetic energy to be efficiently multiplied within the tube. Planck's constant h = 6.63 × 10⁻³⁴ J s, 1 eV = 1.6 × 10⁻¹⁹ J.

Question 1: Which material(s) from the list will respond to light of frequency 5 × 10¹⁴ Hz (photon energy ≈ 2.07 eV)? Justify your answer.

Photon energy at 5 × 10¹⁴ Hz: E = hν = 6.63 × 10⁻³⁴ × 5 × 10¹⁴ ≈ 3.315 × 10⁻¹⁹ J ≈ 2.07 eV. For photoelectric emission, E must exceed the work function φ₀.
Only caesium (φ₀ = 2.14 eV) is close but 2.07 eV < 2.14 eV, so caesium just barely does not emit at this frequency. None of the four materials emit at 5 × 10¹⁴ Hz since all have work functions ≥ 2.14 eV > 2.07 eV. The team needs a material with φ₀ < 2.07 eV.

Question 2: For caesium illuminated by light of frequency 7 × 10¹⁴ Hz (blue-violet light), calculate the maximum kinetic energy of the emitted photoelectrons and explain why caesium is the best choice among the four materials for visible light detection.

E = hν = 6.63 × 10⁻³⁴ × 7 × 10¹⁴ = 4.641 × 10⁻¹⁹ J = 2.90 eV. Kmax = 2.90 − 2.14 = 0.76 eV. The emitted photoelectrons have maximum kinetic energy of 0.76 eV.
Caesium has the lowest work function (2.14 eV) among the four, meaning its threshold frequency is lowest (ν₀ = φ₀/h ≈ 5.16 × 10¹⁴ Hz, in the visible range). The other metals (sodium barely, zinc and platinum not at all) require higher-energy photons. Caesium responds to the widest range of visible frequencies, making it best for detecting faint starlight.

Question 3: If the team doubles the frequency of light used to illuminate the caesium detector (from 7 × 10¹⁴ Hz to 1.4 × 10¹⁵ Hz), will the maximum kinetic energy of photoelectrons double? Show with calculation and explain the physical reason.

At ν = 7 × 10¹⁴ Hz: Kmax = hν − φ₀ = 2.90 − 2.14 = 0.76 eV. At ν' = 1.4 × 10¹⁵ Hz: hν' = 6.63 × 10⁻³⁴ × 1.4 × 10¹⁵ = 9.282 × 10⁻¹⁹ J = 5.80 eV. K'max = 5.80 − 2.14 = 3.66 eV.
The ratio K'max/Kmax = 3.66/0.76 ≈ 4.8, not 2. Doubling frequency does not double Kmax because Kmax = hν − φ₀ is not simply proportional to ν; the work function φ₀ is subtracted as a constant offset.
The physical reason is that part of the photon energy (φ₀ = 2.14 eV) always goes toward overcoming the surface barrier. Only the excess energy (hν − φ₀) appears as kinetic energy. When frequency doubles, the photon energy doubles but the work function subtraction remains constant, so the kinetic energy more than doubles.

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Chapter 11: Dual Nature of Radiation and Matter | Case Study Questions

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Question 1: What is the maximum kinetic energy of the photoelectrons emitted from the caesium plate when light of frequency 8.0 × 10¹⁴ Hz is incident on it? (h = 6.63 × 10⁻³⁴ J s, 1 eV = 1.6 × 10⁻¹⁹ J)

Question 2: When the intensity of the incident light is tripled (frequency kept constant), what happens to the stopping potential and the saturation current? Justify your answer.

Question 3: Explain why classical wave theory fails to account for the observations in this experiment, specifically addressing the existence of a definite stopping potential that is independent of intensity.

Passage

Question 1: Calculate the energy of a photon of UV light with wavelength 200 nm. (h = 6.63 × 10⁻³⁴ J s, c = 3 × 10⁸ m/s, 1 eV = 1.6 × 10⁻¹⁹ J)

Question 2: Which metals will show photoelectric emission under green light (wavelength 550 nm, photon energy ≈ 2.26 eV)? Justify using the concept of threshold frequency.

Question 3: For zinc illuminated by UV light (photon energy 6.22 eV), calculate the maximum kinetic energy of the emitted photoelectrons and the corresponding stopping potential. Explain why increasing the UV intensity will not change the stopping potential.

Passage

Question 1: What does the x-intercept of the stopping potential versus frequency graph represent, and what is the work function of sodium using the data given? (e = 1.6 × 10⁻¹⁹ C)

Question 2: Why is the slope of the V₀ versus frequency graph the same for all metals, while the x-intercept differs from metal to metal?

Question 3: Millikan aimed to disprove Einstein's equation but instead confirmed it. Explain what Einstein's equation predicts about the V₀ versus frequency relationship, and discuss the significance of Millikan's confirmation for the acceptance of the photon model.

Passage

Question 1: Calculate the de Broglie wavelength of the cricket ball (mass 0.15 kg, speed 30 m/s). Why is this wavelength not observable? (h = 6.63 × 10⁻³⁴ J s)

Question 2: An electron is accelerated through a potential difference of 100 V. Calculate its de Broglie wavelength. (m = 9.11 × 10⁻³¹ kg, e = 1.6 × 10⁻¹⁹ C, h = 6.63 × 10⁻³⁴ J s)

Question 3: De Broglie's hypothesis was based on a symmetry argument between radiation and matter. Explain this argument and discuss how the de Broglie relation λ = h/p inherently embodies the dual nature of matter.

Passage

Question 1: Which material(s) from the list will respond to light of frequency 5 × 10¹⁴ Hz (photon energy ≈ 2.07 eV)? Justify your answer.

Question 2: For caesium illuminated by light of frequency 7 × 10¹⁴ Hz (blue-violet light), calculate the maximum kinetic energy of the emitted photoelectrons and explain why caesium is the best choice among the four materials for visible light detection.

Question 3: If the team doubles the frequency of light used to illuminate the caesium detector (from 7 × 10¹⁴ Hz to 1.4 × 10¹⁵ Hz), will the maximum kinetic energy of photoelectrons double? Show with calculation and explain the physical reason.

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