Chapter 5 · Black body radiation & Planck's explanation: practice questions
Unit 2 — Introduction to Modern Physics · Physics. These are 10 questions from the Magmo3 bank on this lesson, from easy to hard. Solve each one, then open "Show answer and explanation". Magmo3 has 45 questions on this lesson, each with an explanation.
A perfectly black body is best described as an ideal body that…
Aemits radiation of one wavelength only
Ballows all the radiation incident upon it to pass through it
Cabsorbs all the electromagnetic radiation incident upon it at every wavelength
Dreflects all the radiation incident upon it
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Correct answer: (C) absorbs all the electromagnetic radiation incident upon it at every wavelength
Explanation: A perfect black body absorbs, and when hot re-emits, all radiation falling on it at every wavelength.
Question 2Easy
For photons of electromagnetic radiation, the energy of a photon is…
Aindependent of its frequency
Binversely proportional to the square of its wavelength
Cdirectly proportional to its frequency
Dinversely proportional to its frequency
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Correct answer: (C) directly proportional to its frequency
Explanation: E = hf, so photon energy is directly proportional to frequency (and inversely proportional to wavelength).
Question 3Easy
Which of the following relations correctly expresses Wien's displacement law for a black body, where λ_max is the wavelength of maximum intensity and T is the absolute temperature?
Aλ_max ∝ T⁴
Bλ_max ∝ 1/T⁴
Cλ_max ∝ T
Dλ_max ∝ 1/T
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Correct answer: (D) λ_max ∝ 1/T
Explanation: Wien's law λ_max × T = constant means λ_max is inversely proportional to the absolute temperature.
Question 4Medium
The surface of a star behaves as a black body at an absolute temperature of 5800 K. Taking Wien's constant b = 2.9 × 10⁻³ m·K, the wavelength at which the emitted intensity is maximum equals…
A quantum of radiation carries an energy of 3.315 × 10⁻¹⁹ J. Taking h = 6.63 × 10⁻³⁴ J·s, the frequency of this radiation equals…
A2.2 × 10⁻¹⁵ Hz
B2.0 × 10¹⁴ Hz
C5.0 × 10¹⁴ Hz
D5.0 × 10⁴⁸ Hz
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Correct answer: (C) 5.0 × 10¹⁴ Hz
Explanation: f = E/h = (3.315 × 10⁻¹⁹)/(6.63 × 10⁻³⁴) = 5.0 × 10¹⁴ Hz.
Question 6Medium
A photon of ultraviolet radiation has a frequency of 1.5 × 10¹⁵ Hz. Taking Planck's constant h = 6.63 × 10⁻³⁴ J·s, the energy of this photon equals…
A9.9 × 10⁻¹⁹ J
B6.6 × 10⁻³⁴ J
C4.4 × 10⁻¹⁹ J
D2.3 × 10¹⁵ J
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Correct answer: (A) 9.9 × 10⁻¹⁹ J
Explanation: E = hf = (6.63 × 10⁻³⁴)(1.5 × 10¹⁵) = 9.9 × 10⁻¹⁹ J.
Question 7Medium
A source emits monochromatic radiation of frequency f. If the source radiates a total energy E, the number of photons (quanta) emitted is given by…
AE·h·f
BE·f/h
CE/(h·f)
Dh·f/E
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Correct answer: (C) E/(h·f)
Explanation: Each photon has energy hf, so the number emitted is the total energy divided by hf, giving n = E/(hf).
Question 8Hard
A photon has a wavelength of 3.0 × 10⁻⁷ m. Taking h = 6.63 × 10⁻³⁴ J·s, c = 3 × 10⁸ m/s and 1 eV = 1.6 × 10⁻¹⁹ J, the energy of the photon in electron-volts is closest to…
A8.3 eV
B4.1 eV
C6.6 eV
D2.1 eV
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Correct answer: (B) 4.1 eV
Explanation: E = hc/λ = 6.63 × 10⁻¹⁹ J; dividing by 1.6 × 10⁻¹⁹ J/eV gives ≈ 4.1 eV.
Question 9Hard
Monochromatic radiation of frequency 5 × 10¹⁴ Hz delivers energy at a rate of 1.989 × 10⁻³ W to a surface. Taking h = 6.63 × 10⁻³⁴ J·s, the number of photons striking the surface each second equals…
A1.5 × 10¹⁶ photons
B3.3 × 10⁻¹⁹ photons
C6.0 × 10¹⁵ photons
D3.0 × 10¹⁵ photons
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Correct answer: (C) 6.0 × 10¹⁵ photons
Explanation: Photon energy = hf = 3.315 × 10⁻¹⁹ J, so n = power/E = (1.989 × 10⁻³)/(3.315 × 10⁻¹⁹) = 6.0 × 10¹⁵ photons per second.
Question 10Hard
Three black bodies A, B and C have their intensity–wavelength peaks at wavelengths λA > λB > λC. Which of the following correctly expresses the relationship between their absolute temperatures?
ATA > TB > TC
BTA = TB = TC
CTB > TA > TC
DTC > TB > TA
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Correct answer: (D) TC > TB > TA
Explanation: By Wien's law T ∝ 1/λ_max, so the body with the shortest peak wavelength (C) is the hottest: TC > TB > TA.