Class 9 Physics Chapter 11 Dual Nature of Radiation and Matter: Important Questions with Complete Solutions

Dual Nature of Radiation and Matter is a pivotal chapter in Class 9 Physics because it shifts student thinking from macroscopic to microscopic phenomena. The photoelectric effect demonstrates that light behaves as packets of energy (photons), overturning the purely wave model. De Broglie's hypothesis extends this duality to matter itself—every particle has an associated wavelength. Together, these concepts underpin modern electronics (photodiodes, image sensors), atomic models, and quantum mechanics. In the revised CBSE assessment pattern, examiners prioritize conceptual understanding over formula-memorization. You'll see questions asking: Why does the photoelectric effect prove light has particle nature? How do we explain electron diffraction? What is the significance of the threshold frequency? These aren't simple plug-and-play problems—they demand reasoning. Chapter 11 carries 8–12% weight in the Class 9 final exam and often appears as 1–2 questions of mixed difficulty. A strong grasp here also builds your foundation for Class 10 and Class 12 physics, where quantum concepts resurface in atomic structure and modern physics.

**Question 1:** In the photoelectric effect, the minimum energy required for an electron to escape from a metal surface is called: (A) Kinetic energy (B) Work function (C) Photon energy (D) Threshold energy **Answer:** (B) Work function. The work function (W or φ) is the minimum energy needed to eject an electron. It depends on the metal's nature. At the threshold frequency (ν₀), photon energy equals work function: hν₀ = W. --- **Question 2:** The stopping potential V_s in a photoelectric experiment is the potential difference that: (A) Accelerates electrons towards the anode (B) Just prevents the fastest photoelectrons from reaching the anode (C) Generates X-rays (D) Ionizes the metal surface **Answer:** (B) Just prevents the fastest photoelectrons from reaching the anode. The kinetic energy of the fastest electron equals eV_s. This relationship, eV_s = ½m_e v_max², is central to Einstein's photoelectric equation. --- **Question 3:** De Broglie wavelength of a particle is given by λ = h/p, where h is Planck's constant and p is: (A) Potential energy (B) Power (C) Momentum (D) Pressure **Answer:** (C) Momentum. De Broglie proposed that all matter has wave properties. The wavelength λ is inversely proportional to momentum p = mv. Higher momentum means shorter wavelength. --- **Question 4:** An electron and a proton have the same de Broglie wavelength. The electron has: (A) Greater kinetic energy (B) Smaller kinetic energy (C) Equal kinetic energy (D) Kinetic energy cannot be compared **Answer:** (A) Greater kinetic energy. Since λ = h/p and wavelengths are equal, momenta are equal. But KE = p²/2m. For the same p, the lighter electron (m_e < m_p) has larger KE. --- **Question 5:** The photoelectric effect is NOT observed when: (A) Light intensity increases (B) Light frequency is less than threshold frequency (C) The metal is exposed to ultraviolet light (D) The anode is positively charged **Answer:** (B) Light frequency is less than threshold frequency. No matter how bright the light, if ν < ν₀, no electrons are ejected. Frequency, not intensity, determines emission. This key observation disproved the classical wave model.

**Question 1:** Explain why the photoelectric effect provides evidence that light has particle nature. **Answer:** The photoelectric effect shows that light transfers energy to electrons in discrete packets (photons), not continuously as classical wave theory predicted. (1) Einstein's photon model: E = hν. If a photon's frequency ν exceeds the threshold ν₀ and satisfies hν ≥ W, an electron is ejected instantly—there is no delay for energy accumulation. (2) Intensity independence: Doubling light intensity doesn't increase the maximum kinetic energy of ejected electrons; it only increases the number of electrons. If light were purely a wave, intense light should transfer more energy per electron. This contradiction proves light is made of energy quanta (photons), each carrying energy hν. Thus, light exhibits particle behaviour in the photoelectric effect. --- **Question 2:** State Einstein's photoelectric equation and define each term. **Answer:** Einstein's photoelectric equation is: **hν = W + KE_max**, where: - **h** = Planck's constant (6.63 × 10⁻³⁴ J·s) - **ν** = frequency of incident light (Hz) - **W** = work function of the metal (J or eV) - **KE_max** = maximum kinetic energy of ejected electrons This equation states that the photon energy (hν) is used partly to overcome the work function W and partly to give kinetic energy to the electron. If hν < W, no emission occurs. --- **Question 3:** What is meant by threshold frequency? Give its relationship with work function. **Answer:** **Threshold frequency (ν₀)** is the minimum frequency of light required to cause photoelectric emission from a metal. Below this frequency, no electrons are ejected, regardless of light intensity. At threshold, the incident photon energy equals the work function: **hν₀ = W**, so **ν₀ = W/h**. This means each metal has its own threshold frequency, determined by its work function. --- **Question 4:** Calculate the de Broglie wavelength of an electron moving at 10% the speed of light. **Answer:** Given: v = 0.1c = 0.1 × 3 × 10⁸ = 3 × 10⁷ m/s; h = 6.63 × 10⁻³⁴ J·s; m_e = 9.11 × 10⁻³¹ kg. Momentum: p = m_e × v = 9.11 × 10⁻³¹ × 3 × 10⁷ = 2.73 × 10⁻²³ kg·m/s De Broglie wavelength: λ = h/p = (6.63 × 10⁻³⁴) / (2.73 × 10⁻²³) = **2.43 × 10⁻¹¹ m = 0.0243 nm** This tiny wavelength explains why electron diffraction was difficult to observe until Davisson and Germer's experiments. --- **Question 5:** Distinguish between photons and photoelectrons. **Answer:** **Photons** are quanta (packets) of electromagnetic energy with energy E = hν and momentum p = h/λ. They have no rest mass and always travel at the speed of light. **Photoelectrons** are electrons ejected from a metal surface due to incident photon absorption. They have rest mass (m_e = 9.11 × 10⁻³¹ kg) and move with velocities much less than c. Photons transfer their energy to electrons, creating photoelectrons.

**Question 1:** A metal has a work function of 3.2 eV. Will it show photoelectric effect when exposed to green light (λ = 500 nm)? Calculate the stopping potential if the answer is yes. **Solution:** Step 1: Calculate photon energy. E = hc/λ = (6.63 × 10⁻³⁴ × 3 × 10⁸) / (500 × 10⁻⁹) = (19.89 × 10⁻²⁶) / (5 × 10⁻⁷) = 3.98 × 10⁻¹⁹ J Convert to eV: E = (3.98 × 10⁻¹⁹) / (1.6 × 10⁻¹⁹) = 2.49 eV Step 2: Compare with work function. Since 2.49 eV < 3.2 eV, **photoelectric effect will NOT occur**. The photon energy is insufficient to overcome the work function. No electrons will be ejected. --- **Question 2:** Explain the wave nature of matter using de Broglie's hypothesis. Why is the wave nature of large objects not observable? **Solution:** De Broglie's hypothesis (1924): Every moving particle has an associated wavelength λ = h/p. Just as light exhibits wave-particle duality, matter exhibits the same duality. An electron or atom in motion behaves like a wave; this was later confirmed by electron diffraction (Davisson–Germer experiment). **Why large objects don't show wave behaviour:** For a macroscopic object (e.g., a 1 kg ball moving at 10 m/s): p = mv = 1 × 10 = 10 kg·m/s λ = h/p = (6.63 × 10⁻³⁴) / 10 = 6.63 × 10⁻³⁵ m This wavelength is ~10²⁰ times smaller than an atom. Diffraction and interference effects require the wavelength to be comparable to obstacles or slits (on the order of 10⁻⁶ m or larger). Since 6.63 × 10⁻³⁵ m is impossibly small, we never observe wave effects in large objects. Thus, quantum effects are negligible at macroscopic scales. --- **Question 3:** Photoelectrons are ejected from a zinc surface with a stopping potential of 1.5 V. Calculate the work function of zinc if the incident light has a frequency of 1.5 × 10¹⁵ Hz. **Solution:** Given: V_s = 1.5 V; ν = 1.5 × 10¹⁵ Hz; h = 6.63 × 10⁻³⁴ J·s; 1 eV = 1.6 × 10⁻¹⁹ J Step 1: Maximum kinetic energy of ejected electrons. KE_max = eV_s = 1.6 × 10⁻¹⁹ × 1.5 = 2.4 × 10⁻¹⁹ J Step 2: Photon energy. hν = 6.63 × 10⁻³⁴ × 1.5 × 10¹⁵ = 9.945 × 10⁻¹⁹ J Step 3: Work function (from Einstein's equation: hν = W + KE_max). W = hν − KE_max = 9.945 × 10⁻¹⁹ − 2.4 × 10⁻¹⁹ = 7.545 × 10⁻¹⁹ J Convert to eV: W = (7.545 × 10⁻¹⁹) / (1.6 × 10⁻¹⁹) = **4.72 eV** --- **Question 4:** Two particles, A (electron) and B (neutron), have the same de Broglie wavelength. Which has greater momentum and kinetic energy? (m_e = 9.11 × 10⁻³¹ kg; m_n ≈ 1.67 × 10⁻²⁷ kg) **Solution:** Given: λ_A = λ_B, so p_A = p_B (since λ = h/p and h is constant). **Momentum:** Both have equal momentum p. **Kinetic Energy:** KE = p²/(2m) For particle A (electron): KE_A = p²/(2m_e) For particle B (neutron): KE_B = p²/(2m_n) Since m_e < m_n, we have **KE_A > KE_B**. **Conclusion:** Both particles have equal momentum. The electron (lighter particle) has greater kinetic energy. This illustrates the inverse relationship between mass and kinetic energy at constant momentum.

**Question 1:** State and explain the photoelectric effect. Discuss how Einstein's photon model resolves the shortcomings of classical wave theory. **Complete Solution:** **Statement of Photoelectric Effect:** When light of sufficiently high frequency is incident on a metal surface, electrons are emitted from the surface. These emitted electrons are called photoelectrons. The phenomenon is called the photoelectric effect. **Classical Wave Theory's Predictions and Failures:** Classical electromagnetic theory treated light as a continuous wave, predicting that: 1. Higher intensity light should produce faster electrons (more energy accumulation over time). 2. Very low-frequency light, if intense enough, should eventually cause emission (given enough time for energy to accumulate). 3. There should be a time delay between light incidence and electron emission. Experimental observations contradicted all three predictions: - Maximum KE of electrons depended only on light frequency, not intensity. - No emission below a threshold frequency, regardless of intensity or duration. - Emission was instantaneous (< 10⁻⁹ s), with no observable delay. **Einstein's Photon Model (1905):** Einstein proposed that light consists of discrete packets (quanta) of energy called photons, each carrying E = hν. When a photon strikes a metal, it transfers all its energy to a single electron in a single collision—not gradually as a wave. **Einstein's Photoelectric Equation:** hν = W + KE_max where hν is photon energy, W is work function, and KE_max is maximum kinetic energy of the ejected electron. **How Einstein's Model Resolves the Contradictions:** 1. **Frequency dependence:** High-frequency photons carry more energy (E = hν). An electron gains energy from a single photon, so maximum KE = hν − W depends only on ν, not intensity. ✓ 2. **Threshold frequency:** Below ν₀ where hν₀ = W, no photon carries enough energy to overcome the work function. Intensity is irrelevant because intensity only increases the number of photons, not their individual energy. ✓ 3. **No time delay:** Each photon–electron collision is instantaneous; there is no accumulation phase. Energy transfer is particle-like, not wavelike. ✓ **Key Evidence from Experiments:** - Stopping potential V_s increases linearly with frequency: eV_s = h(ν − ν₀), confirming the linear relationship in Einstein's equation. - Slope of V_s vs. ν graph gives h/e, allowing measurement of Planck's constant. - Work function varies with metal, matching observations. **Conclusion:** Einstein's photon model treats light as discrete particles (quanta), explaining why light exhibits particle behaviour in the photoelectric effect. This was revolutionary: light, traditionally viewed as a wave, has a dual nature—it behaves as both a wave and a particle depending on the context. --- **Question 2:** Derive the de Broglie wavelength formula and explain its significance in atomic physics. How does electron diffraction prove wave nature of matter? **Complete Solution:** **Derivation of de Broglie Wavelength:** Step 1: Light has dual nature (confirmed by photoelectric effect). - As a wave: c = νλ, so λ = c/ν = c/(E/h) = hc/E - As a particle (photon): E = hν and momentum p = E/c - Combining: p = E/c and λ = c/ν = c/(E/h) - Therefore: λ = h/p Step 2: De Broglie's hypothesis (1924): Extend wave-particle duality to matter. If matter has both particle and wave properties, and if the relationship λ = h/p holds for photons, then the same relationship should apply to all moving particles: **λ = h/p = h/(mv)** where m is mass, v is velocity, h is Planck's constant, and p = mv is momentum. **Significance in Atomic Physics:** 1. **Explains atomic stability:** Classical physics predicted that electrons spiraling into the nucleus should emit radiation continuously and collapse. De Broglie's hypothesis suggests electrons have wave properties; stable orbits correspond to standing wave patterns where the circumference equals an integer multiple of wavelengths: 2πr = nλ. This naturally explains quantized orbits without additional postulates. 2. **Foundation of quantum mechanics:** The wave nature of matter led to the Schrödinger wave equation, which describes electron probability distributions (orbitals) rather than definite trajectories. This resolves the classical contradictions of atomic structure. 3. **Explains electron diffraction:** Electrons behave like waves when passing through narrow slits or crystal lattices, producing interference and diffraction patterns just like light. **Electron Diffraction (Davisson–Germer Experiment, 1927):** **Experimental Setup:** A beam of electrons accelerated through a potential V is directed at a nickel crystal. The electrons are scattered, and the intensity of scattered electrons is measured as a function of scattering angle. **Theory:** When electrons pass through the crystal lattice (spacing d ≈ 0.1 nm), they are scattered by atoms. If electrons have wavelength λ = h/p, constructive interference occurs when the path difference equals an integer multiple of λ: 2d sinθ = nλ (Bragg's law) This produces maxima (bright spots) and minima (dark spots) in the scattered electron pattern, exactly like X-ray diffraction. **Experimental Observation:** Davisson and Germer found strong diffraction maxima at specific angles, consistent with Bragg's law using λ = h/p. For electrons accelerated through V = 54 V: p = √(2m_e eV) = √(2 × 9.11 × 10⁻³¹ × 1.6 × 10⁻¹⁹ × 54) ≈ 4.1 × 10⁻²⁴ kg·m/s λ = h/p ≈ 0.165 nm This matched the observed diffraction pattern, confirming wave nature of electrons. **Proof of Matter's Wave Nature:** Electron diffraction demonstrates that matter exhibits wave properties (interference, diffraction) under suitable experimental conditions, just as light does in the double-slit experiment. This conclusively proved de Broglie's hypothesis and established the wave-particle duality of matter—a cornerstone of quantum mechanics. --- **Question 3:** A light source of frequency 5 × 10¹⁴ Hz illuminates a copper surface (work function = 4.4 eV). Calculate: (a) photon energy in eV; (b) kinetic energy of ejected electrons; (c) stopping potential; (d) de Broglie wavelength of ejected electrons. [h = 6.63 × 10⁻³⁴ J·s; 1 eV = 1.6 × 10⁻¹⁹ J; m_e = 9.11 × 10⁻³¹ kg; c = 3 × 10⁸ m/s] **Complete Solution:** **(a) Photon Energy in eV:** E_photon = hν = 6.63 × 10⁻³⁴ × 5 × 10¹⁴ = 3.315 × 10⁻¹⁹ J Convert to eV: E = (3.315 × 10⁻¹⁹) / (1.6 × 10⁻¹⁹) = **2.07 eV** **(b) Kinetic Energy of Ejected Electrons:** Using Einstein's equation: hν = W + KE_max KE_max = hν − W = 2.07 − 4.4 = **−2.33 eV** **Result:** KE_max is negative, meaning **no photoelectric effect occurs**. The photon energy (2.07 eV) is less than the work function (4.4 eV), so electrons cannot be ejected. **Verification:** Threshold frequency ν₀ = W/h = 4.4 eV / (6.63 × 10⁻³⁴ J·s) = 4.4 × 1.6 × 10⁻¹⁹ / (6.63 × 10⁻³⁴) = 1.06 × 10¹⁵ Hz. Since the incident frequency (5 × 10¹⁴ Hz) < threshold frequency, no emission occurs. ✓ **(c) Stopping Potential:** Since no electrons are ejected, **stopping potential = 0 V** (or undefined). There is no stopping potential because there are no photoelectrons to stop. **(d) De Broglie Wavelength:** Again, since no photoelectrons are produced, **de Broglie wavelength is not applicable**. **Conclusion:** For photoelectric emission from copper to occur, the incident light frequency must exceed 1.06 × 10¹⁵ Hz (or wavelength < 283 nm, in the ultraviolet range). Green, yellow, or red light cannot cause emission from copper.

**Case Study: Photoelectron Imaging in Smartphones** Modern smartphone cameras and image sensors rely on the photoelectric effect to convert light into electrical signals. When a photon strikes a pixel (made of a light-sensitive semiconductor), it ejects electrons that are collected to create an image. Consider a smartphone camera with pixels sensitive to green light (λ = 550 nm). The semiconductor material has a work function of 1.8 eV. **(a) Will photoelectric emission occur when the camera is exposed to green light? Justify numerically.** **Solution:** Photon energy: E = hc/λ = (6.63 × 10⁻³⁴ × 3 × 10⁸) / (550 × 10⁻⁹) = 3.6 × 10⁻¹⁹ J Convert to eV: E = (3.6 × 10⁻¹⁹) / (1.6 × 10⁻¹⁹) = **2.25 eV** Since 2.25 eV > 1.8 eV (work function), **yes, photoelectric emission occurs**. The photon energy exceeds the work function, so electrons are ejected. --- **(b) Calculate the maximum kinetic energy of ejected electrons and the stopping potential.** **Solution:** Maximum KE: KE_max = hν − W = 2.25 − 1.8 = **0.45 eV** Stopping potential: eV_s = KE_max V_s = 0.45 V = **0.45 V** This means the fastest photoelectrons can overcome a retarding potential of 0.45 V before stopping. --- **(c) If the smartphone camera detects red light (λ = 700 nm) instead, explain why the image quality might differ, even if photoelectric emission still occurs.** **Solution:** Photon energy for red light: E_red = hc/700 × 10⁻⁹ = (6.63 × 10⁻³⁴ × 3 × 10⁸) / (700 × 10⁻⁹) = 2.84 × 10⁻¹⁹ J ≈ 1.78 eV Since 1.78 eV < 1.8 eV, **red light photons have energy just below (or equal to, near the edge) the work function**. Few or no electrons are ejected, or they're ejected with near-zero kinetic energy. **Why image quality differs:** - Fewer photoelectrons are collected per pixel → lower signal → darker, grainier image. - Signal-to-noise ratio decreases, reducing image clarity and contrast. - This is why smartphone cameras perform poorly in red light or low light; sensitivity drops below the work function threshold. --- **(d) To improve sensitivity to infrared light (λ = 1000 nm), the camera manufacturer wants to reduce the work function. What work function would allow photoelectric emission from infrared photons? Suggest a practical material.** **Solution:** For infrared (λ = 1000 nm): E_IR = hc/λ = (6.63 × 10⁻³⁴ × 3 × 10⁸) / (1000 × 10⁻⁹) = 1.99 × 10⁻¹⁹ J ≈ 1.24 eV For photoelectric emission: W ≤ E_IR **W ≤ 1.24 eV** Practical materials: - **Cesium** (Cs): W ≈ 2.0 eV — marginal for IR - **Sodium** (Na): W ≈ 2.28 eV — too high - **Potassium** (K): W ≈ 2.3 eV — too high - **Selenium** (Se): W ≈ 5.11 eV — too high for IR - **Silicon** (Si): W ≈ 4.83 eV — suitable for visible, not IR **Practical approach:** Use a narrow-bandgap **InGaAs (Indium Gallium Arsenide)** semiconductor, which has an effective work function around 0.4–0.7 eV and is highly sensitive to near-infrared (800–1700 nm). This is why professional IR cameras and night-vision sensors use InGaAs detectors. --- **(e) Using de Broglie's concept, explain why increasing the retarding potential (stopping potential) reduces the number of ejected electrons collected at the anode.** **Solution:** When stopping potential V_s is applied, it creates a retarding electric field opposing the motion of photoelectrons. For an electron with initial kinetic energy KE: - It can overcome the retarding field only if KE > eV_s. - Those with KE ≤ eV_s are turned back before reaching the anode. From de Broglie's perspective: - Electrons with higher KE have shorter wavelengths (λ = h/p, and p ∝ √KE). - These high-KE electrons penetrate further against the retarding field. - As V_s increases, only electrons with λ below a critical threshold can reach the anode. - The number of electrons with sufficient momentum (short enough wavelength) decreases. At V_s = stopping potential (calculated in part b as 0.45 V), even the fastest electrons cannot overcome the field, and **photocurrent drops to zero**. This is how stopping potential is experimentally measured: sweep V_s upward until current reaches zero. **Conclusion:** The stopping potential technique exploits both the energy and wave properties of photoelectrons to characterize the photoelectric effect comprehensively.

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