Free Dawn of Modern Physics MCQs with Answers
739 Dawn of Modern Physics MCQs from Physics, each with the correct answer and a written explanation of why it is correct. Free and unlimited, with no account needed.
Classical physics cannot explain blackbody radiation and related observations, leading to Planck’s quantum theory, in which energy is emitted or absorbed in discrete packets. Photons provide the particle model of light, with energy and momentum linked to frequency and wavelength, including the photoelectric effect and its threshold frequency.
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Read the Dawn of Modern Physics notesFree MDCAT chapter notes with key terms739 questions · page 13 of 37
- A. 3 × 10¹⁸ kg/sec
- B. 3 × 10¹⁹ kg/sec
- C. 3 × 10²⁰ kg/sec
- D. 3 × 10²¹ kg/sec
Explanation: Using E = mc², mass loss m = E/c² = (2.7 × 10³⁶)/(9 × 10¹⁶) = 3 × 10¹⁹ kg/sec.
Correct answer: 3 × 10¹⁹ kg/sec242. When photons of energy hv are incident on a photosensitive material of work function hv₀, then:
- A. The kinetic energy of all emitted electrons is hv₀
- B. The kinetic energy of all emitted electrons is h(v-v₀)
- C. The kinetic energy of the fastest electron is h(v-v₀)
- D. The kinetic energy of all emitted electrons is hv
Explanation: Photon energy is hv, work function is hv₀. Maximum kinetic energy of emitted electrons is h(v - v₀).
Correct answer: The kinetic energy of the fastest electron is h(v-v₀)- A. 3.09 eV
- B. 1.41 eV
- C. 1.51 eV
- D. 1.68 eV
Explanation: Photon energy E = hc/λ = 1240/400 = 3.1 eV. Kinetic energy KE = 1.68 eV. Work function Φ = E - KE = 3.1 - 1.68 = 1.42 eV ≈ 1.41 eV.
Correct answer: 1.41 eV- A. Are one quarter as numerous
- B. Are half as numerous
- C. Each carry one quarter of their previous momentum
- D. Each carry one quarter of their previous energy
Explanation: Intensity decreases with distance squared, I' = I/(2)² = I/4. Photoelectron number is proportional to intensity, reducing to one-quarter.
Correct answer: Are one quarter as numerous- A. 1.2 eV
- B. 1.5 eV
- C. 1.6 eV
- D. 1.8 eV
Explanation: Threshold energy Φ = hc/λ₀ = 1240/2300 ≈ 5.39 eV. Photon energy E = hc/λ = 1240/1800 ≈ 6.89 eV. Maximum KE = E - Φ = 6.89 - 5.39 ≈ 1.5 eV.
Correct answer: 1.5 eV- A. 2 times the earlier value
- B. More than 2 times the earlier value
- C. Less than 2 times the earlier value
- D. Unchanged
Explanation: Initial KE = h(v - v₀). Doubled frequency: KE₁ = h(2v - v₀) = 2h(v - v₀) + hv₀ = 2KE + hv₀, which is more than doubled due to the hv₀ term.
Correct answer: More than 2 times the earlier value- A. 54 keV
- B. 76 keV
- C. 88 keV
- D. 32 keV
Explanation: The K line energy corresponds to the difference in stopping potentials, calculated as 76 keV based on the energy transitions.
Correct answer: 76 keV- A. 0.48 μA
- B. 0.42 mA
- C. 0.48 mA
- D. 0.42 μA
Explanation: Photon energy E = hc/λ = 1240/400 ≈ 3.1 eV = 4.96 × 10⁻¹⁹ J. Number of photons n = P/E = (1.5 × 10⁻³)/(4.96 × 10⁻¹⁹) ≈ 3.02 × 10¹⁵.
Correct answer: 0.48 μA- A. 4 eV
- B. 6.2 eV
- C. 2 eV
- D. 2.2 eV
Explanation: Stopping potential energy = qEd = 1.6 × 10⁻¹⁹ × 4 × 1 ≈ 4 eV. Photon energy E = hc/λ = 1240/200 = 6.2 eV.
Correct answer: 2.2 eV- A. 309 nm
- B. 209 nm
- C. 109 nm
- D. 9 nm
Explanation: Threshold wavelength λ = hc/Φ = 1240/4 ≈ 310 nm ≈ 309 nm.
Correct answer: 309 nm- A. 6.2 × 10²⁰
- B. 3 × 10¹⁹
- C. 1.5 × 10²⁰
- D. 6 × 10¹⁸
Explanation: Photon energy E = hc/λ = (6.6 × 10⁻³⁴ × 3 × 10⁸)/(0.6 × 10⁻⁶) ≈ 3.3 × 10⁻¹⁹ J. Effective power = 0.25 × 200 = 50 W.
Correct answer: 1.5 × 10²⁰- A. 50
- B. 5
- C. 500
- D. More than 5 million
Explanation: Energy of one photon at 6000 Å =3.3 ✕ 10^(-19)J.power delivered =1.7 ✕10^ (-18) J/s, so photon per second =1.7✕ 10^(-18) /1.3 ✕ 10^(-19)=5…
Correct answer: More than 5 million- A. Yes
- B. Yes, if the red beam is sufficiently intense
- C. Yes, if the red beam is allowed to fall for sufficient duration
- D. No
Explanation: Red light has less energy than yellow light, which itself is below the work function. So red light cannot cause photoemission.
Correct answer: No- A. Parabola
- B. Straight line
- C. Hyperbola
- D. Circle
Explanation: Stopping potential Vs which is linearly proportional to the frequency above the threshold. So the graph is a straight line.
Correct answer: Straight line- A. 4 times the original current
- B. 2 times the original current
- C. Half the original current
- D. Zero times the original current
Explanation: Original frequency = 1.5v₀. Halving gives 0.75v₀, which is below the threshold frequency v₀, so no photoelectrons are emitted, regardless…
Correct answer: Zero times the original current- A. 2.35 × 10⁻⁴ N
- B. 108 N
- C. 8.35 × 10⁴ N
- D. 8.8 × 10⁻⁸ N
Explanation: Force = I × A/c. Area A = πr² = π × 4 ≈ 12.56 m². Force = (1.4 × 10³ × 12.56)/(3 × 10⁸) ≈ 2.35 × 10⁻⁴ N.
Correct answer: 2.35 × 10⁻⁴ N- A. hv/c²
- B. hv/c
- C. hv
- D. Zero
Explanation: Photons travel at the speed of light and have zero rest mass, as per special relativity.
Correct answer: Zero- A. A
- B. B
- C. C
- D. D
Explanation: The inversely proportional graph correctly represents the variation of particle momentum with associated de Broglie wavelength.
Correct answer: D- A. X-rays
- B. Light waves
- C. y -rays
- D. Sound waves
Explanation: Electromagnetic Waves do not need a medium for their propagation; they can travel in a vacuum.
Correct answer: Sound waves- A. Maxwell
- B. de-Broglie
- C. Newton
- D. Galileo
Explanation: Maxwell was a Scottish physicist who was regarded as being the scientist to regard light as electromagnetic waves.
Correct answer: Maxwell