Dual Nature of Radiation and Matter
Tamil Nadu Board · Class 12 · Physics
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Light of frequency 6 × 10¹⁴ Hz is incident on a metal surface with work function 2.0 eV. What is the maximum kinetic energy of the emitted photoelectrons? (h = 6.626 × 10⁻³⁴ Js, 1 eV = 1.6 × 10⁻¹⁹ J)
In a photoelectric effect experiment, the stopping potential for a metal is found to be 1.5 V. What is the maximum speed of the emitted photoelectrons? (e = 1.6 × 10⁻¹⁹ C, m_e = 9.1 × 10⁻³¹ kg)
Which of the following correctly describes the relationship between the intensity of incident light and photoelectric current?
An electron is accelerated through a potential difference of 100 V. What is the de Broglie wavelength associated with it?
Sample Questions
The work function of cesium is 2.14 eV. What is the threshold frequency of light required for photoelectric emission from cesium? (h = 6.626 × 10⁻³⁴ Js, 1 eV = 1.6 × 10⁻¹⁹ J)
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5.16 × 10¹⁴ Hz
Step 1: At threshold frequency, the photon energy exactly equals the work function. hν₀ = φ₀ Step 2: Convert work function to joules. φ₀ = 2.14 × 1.6 × 10⁻¹⁹ = 3.424 × 10⁻¹⁹ J Step 3: Solve for threshold frequency. ν₀ = φ₀/h = 3.424 × 10⁻¹⁹ / 6.626 × 10⁻³⁴ Step 4: Calculate. ν₀ = 5.16 × 10¹⁴ Hz Why wrong options are wrong: - 3.24 × 10¹⁴ Hz: Results from forgetting to convert eV to joules. - 8.20 × 10¹⁴ Hz: Results from using incorrect conversion factor. - 2.58 × 10¹⁴ Hz: Results from dividing by 2h instead of h.
In the Davisson-Germer experiment, electrons were accelerated through 54 V and directed at a nickel crystal. A diffraction peak was observed at 50° to the incident beam. What does this experiment conclusively prove?
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Electrons exhibit wave nature and undergo diffraction similar to X-rays.
Step 1: In the Davisson-Germer experiment, an electron beam was directed at a nickel crystal and the scattered intensity was measured at different angles. Step 2: A distinct peak (maximum) in intensity was observed at 50°, indicating constructive interference — a characteristic property of waves. Step 3: The experimentally calculated wavelength (1.65 Å) matched the de Broglie wavelength (1.67 Å) calculated for 54 V electrons. Step 4: Diffraction is a wave phenomenon. The fact that electrons showed diffraction conclusively proves the wave nature of matter, confirming de Broglie's hypothesis.
A photon has a wavelength of 500 nm. What is its momentum? (h = 6.626 × 10⁻³⁴ Js)
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1.325 × 10⁻²⁷ kg m/s
Step 1: Use the photon momentum formula. p = h/λ Step 2: Convert wavelength to metres. λ = 500 nm = 500 × 10⁻⁹ m = 5 × 10⁻⁷ m Step 3: Calculate momentum. p = 6.626 × 10⁻³⁴ / 5 × 10⁻⁷ Step 4: Result. p = 1.325 × 10⁻²⁷ kg m/s Why wrong options are wrong: - 2.65 × 10⁻²⁷ kg m/s: Results from using λ = 250 nm (half the given value). - 6.626 × 10⁻³⁴ kg m/s: This is Planck's constant itself, not the momentum. - 3.97 × 10⁻¹⁹ kg m/s: This is the photon energy in joules, not momentum.
Which of the following statements correctly explains why the wave theory of light FAILS to explain the photoelectric effect?
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Wave theory predicts a large time lag before emission, but experiments show instantaneous emission (less than 10⁻⁹ s).
Step 1: According to wave theory, energy of light is spread uniformly over the entire wavefront. Step 2: Each electron absorbs energy continuously and slowly. Calculations show it would take approximately 79 days (for cesium) before an electron accumulates enough energy to overcome the work function. Step 3: But experiments show photoelectric emission is almost instantaneous — within 10⁻⁹ s of light being incident. Step 4: This huge contradiction (days vs nanoseconds) is one of the key failures of wave theory. Wave theory also fails on threshold frequency (it predicts emission at any freque
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