College Physics Quiz: Blackbody Radiation

10 questions · exam conditions

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Blackbody RadiationQuestion 1 of 10

A tungsten filament light bulb operates at 3000 K. If an LED produces the same visible light output but its junction operates at 350 K, what is the ratio of infrared power radiated by the tungsten filament to that radiated by the LED junction, assuming both can be modeled as blackbodies with the same surface area?

8.6

540

4600

39000

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College Physics Quiz

College Physics Quiz: Blackbody Radiation

Practice Blackbody Radiation in College Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Blackbody Radiation, giving you a quick way to practice the rules, question types, and explanations that matter most for College Physics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

8.6
74
540
4600 (correct answer)
39000

Explanation: When you encounter thermal radiation problems comparing different temperatures, you're dealing with blackbody radiation and the Stefan-Boltzmann law. The key insight is that radiated power depends on the fourth power of temperature. The Stefan-Boltzmann law states that power radiated per unit area is

P = \sigma T^4

, where

\sigma

is the Stefan-Boltzmann constant and

T

is absolute temperature. Since both objects have the same surface area, the ratio of their radiated powers equals the ratio of

T^4

values. For the tungsten filament at 3000 K versus the LED at 350 K:

\frac{P_{tungsten}}{P_{LED}} = \frac{(3000)^4}{(350)^4} = \frac{8.1 \times 10^{13}}{1.5 \times 10^{10}} ≈ 4600

This confirms answer D is correct. Answer A (8.6) likely comes from taking the simple ratio of temperatures:

\frac{3000}{350} ≈ 8.6

, but this ignores the fourth-power relationship. Answer B (74) might result from using a squared relationship:

(\frac{3000}{350})^2 ≈ 74

, which is still incorrect. Answer C (540) could come from a cubed relationship or other calculation error, missing the correct fourth-power dependence. Remember that thermal radiation problems almost always involve the fourth power of temperature. When you see blackbody radiation questions, immediately think

T^4

relationship – this is one of the most important scaling laws in thermal physics and appears frequently on physics exams.

Question 2

A blackbody cavity has a small hole through which radiation escapes. If the cavity temperature increases from $T$ to $1.5T$ , by what factor does the power escaping through the hole increase?

1.5
2.25
3.38
5.06 (correct answer)
7.59

Explanation: When you encounter blackbody radiation problems, you're dealing with the Stefan-Boltzmann law, which describes how much power a blackbody radiates. The key relationship is that radiated power is proportional to the fourth power of temperature:

P \propto T^4

. For a blackbody cavity with a hole, the power escaping through that hole follows this same temperature dependence. If the temperature increases from

T

1.5T

, you need to find the ratio of the new power to the original power:

\frac{P_{new}}{P_{original}} = \frac{(1.5T)^4}{T^4} = (1.5)^4 = 5.0625 \approx 5.06

This confirms answer D is correct. The wrong answers represent common mathematical errors. Choice A (1.5) assumes power is simply proportional to temperature—this would be true if you forgot about the fourth-power relationship entirely. Choice B (2.25) comes from incorrectly using a second-power relationship:

(1.5)^2 = 2.25

, which might arise if you confused this with other physics laws that have quadratic dependence. Choice C (3.38) results from using a third-power relationship:

(1.5)^3 = 3.375

, suggesting confusion about the exact exponent in the Stefan-Boltzmann law. Remember: blackbody radiation problems always involve the fourth power of temperature. This

T^4

dependence means small temperature changes produce dramatic changes in radiated power—a doubling of temperature increases power by a factor of 16, making thermal radiation extremely sensitive to temperature variations.

Question 3

Planck's law for blackbody radiation can be written as $B(\lambda,T) = \frac{2hc^2}{\lambda^5} \frac{1}{e^{hc/\lambda kT} - 1}$ . In the limit where $hc/\lambda kT \ll 1$ , this expression reduces to which classical law?

Wien's displacement law
Stefan-Boltzmann law
Rayleigh-Jeans law (correct answer)
Beer's law
Lambert's law

Explanation: When you encounter Planck's law and are asked about limiting cases, you're dealing with the historical transition from classical to quantum physics. The key is recognizing what mathematical approximation to make when

hc/\lambda kT \ll 1

. In this limit, the exponential term

e^{hc/\lambda kT}

is close to 1, so you can use the approximation

e^x \approx 1 + x

for small

x

. This gives us:

e^{hc/\lambda kT} - 1 \approx (1 + \frac{hc}{\lambda kT}) - 1 = \frac{hc}{\lambda kT}

Substituting this back into Planck's law:

B(\lambda,T) = \frac{2hc^2}{\lambda^5} \frac{1}{\frac{hc}{\lambda kT}} = \frac{2hc^2}{\lambda^5} \cdot \frac{\lambda kT}{hc} = \frac{2ckT}{\lambda^4}

This is exactly the Rayleigh-Jeans law, making C correct. A) Wien's displacement law relates the peak wavelength to temperature (

\lambda_{max} \propto 1/T

), not the spectral radiance formula we derived. B) The Stefan-Boltzmann law gives total power radiated (

\propto T^4

) by integrating over all wavelengths, not the wavelength-dependent distribution. D) Beer's law describes light absorption through materials, completely unrelated to blackbody radiation. Remember this pattern: when you see "classical limit" of a quantum equation, look for where

h

(Planck's constant) effectively cancels out or becomes negligible. The resulting expression should match a pre-quantum classical theory.

Question 4

Two blackbody spheres A and B have the same total radiated power. Sphere A has temperature 4000 K and radius 1.0 m. If sphere B has temperature 2000 K, what is its radius?

1.0 m
2.0 m
4.0 m (correct answer)
8.0 m
16.0 m

Explanation: When you encounter blackbody radiation problems involving two objects with equal power output, you're dealing with the Stefan-Boltzmann law, which states that radiated power depends on both temperature and surface area. The Stefan-Boltzmann law gives us:

P = \sigma A T^4

, where

P

is power,

A

is surface area, and

T

is temperature. For spheres,

A = 4\pi r^2

. Since both spheres have equal power output:

P_A = P_B

\sigma (4\pi r_A^2) T_A^4 = \sigma (4\pi r_B^2) T_B^4

Simplifying:

r_A^2 T_A^4 = r_B^2 T_B^4

Solving for

r_B

r_B = r_A \sqrt{\frac{T_A^4}{T_B^4}} = r_A \left(\frac{T_A}{T_B}\right)^2

Substituting values:

r_B = 1.0 \times \left(\frac{4000}{2000}\right)^2 = 1.0 \times 2^2 = 4.0 \text{ m}

Choice A (1.0 m) incorrectly assumes radius stays constant when temperature changes. Choice B (2.0 m) results from using a linear relationship (

T_A/T_B = 2

) instead of the fourth-power relationship. Choice D (8.0 m) comes from cubing the temperature ratio instead of squaring it. Remember that blackbody power scales with

T^4

, so when temperature decreases, surface area must increase dramatically to maintain constant power output. The key relationship is

r \propto T^{-2}

for constant power—when temperature halves, radius must quadruple.

Question 5

The cosmic microwave background radiation is nearly a perfect blackbody with temperature 2.7 K. At what wavelength does this radiation have its peak intensity? (Use $b = 2.90 \times 10^{-3}\text{ m·K}$ )

0.1 mm
1.1 mm (correct answer)
2.7 mm
7.8 mm
29 mm

Explanation: When you encounter questions about blackbody radiation and peak wavelength, you're dealing with Wien's displacement law, which relates the temperature of a blackbody to the wavelength at which it emits most intensely. Wien's displacement law states that

\lambda_{max} = \frac{b}{T}

, where

b = 2.90 \times 10^{-3} \text{ m·K}

is Wien's displacement constant and

T

is the absolute temperature. For the cosmic microwave background at 2.7 K:

\lambda_{max} = \frac{2.90 \times 10^{-3} \text{ m·K}}{2.7 \text{ K}} = 1.07 \times 10^{-3} \text{ m} = 1.07 \text{ mm}

This confirms answer choice B (1.1 mm) as correct. Let's examine why the other options are wrong: A) 0.1 mm would correspond to a much hotter blackbody (29 K), nearly 11 times warmer than the CMB. C) 2.7 mm represents a common trap where students might confuse the temperature value (2.7 K) with the wavelength, but these have different units and aren't equal. D) 7.8 mm would correspond to a cooler blackbody (about 0.37 K), roughly seven times colder than the actual CMB temperature. Remember Wien's law shows an inverse relationship: hotter objects peak at shorter wavelengths (blueshifted), while cooler objects peak at longer wavelengths (redshifted). For blackbody radiation problems, always use Wien's displacement law with the given constant, and watch out for answer choices that simply restate the given temperature value.

Question 6

An electric heating element can be modeled as a blackbody. When connected to 120 V, it draws 10 A and reaches 1200 K. If connected to 240 V, it draws 20 A. Assuming the resistance remains constant, what temperature does it reach at the higher voltage?

1440 K
1697 K (correct answer)
2400 K
4800 K
9600 K

Explanation: This problem combines electrical circuits with blackbody radiation, testing your understanding of how power dissipation relates to temperature through Stefan-Boltzmann radiation. Start by finding the resistance using Ohm's law:

R = V/I = 120V/10A = 12Ω

. Since resistance remains constant, you can verify this at 240V:

R = 240V/20A = 12Ω

. Next, calculate the power dissipated at each voltage. At 120V:

P_1 = VI = 120V × 10A = 1200W

. At 240V:

P_2 = VI = 240V × 20A = 4800W

. Notice that doubling the voltage quadruples the power (since

P = V^2/R

). For a blackbody, the Stefan-Boltzmann law states that radiated power is proportional to

T^4

P = σAT^4

. Since the heating element's surface area remains constant,

P_1/P_2 = T_1^4/T_2^4

. Setting up the ratio:

\frac{1200W}{4800W} = \frac{(1200K)^4}{T_2^4}

Solving:

\frac{1}{4} = \frac{(1200)^4}{T_2^4}

, so

T_2^4 = 4 × (1200)^4

Taking the fourth root:

T_2 = 1200 × 4^{1/4} = 1200 × \sqrt{2} ≈ 1697K

Choice A (1440K) incorrectly assumes linear temperature scaling with voltage. Choice C (2400K) assumes direct proportionality with voltage. Choice D (4800K) mistakenly uses the power ratio as the temperature ratio. Remember: when dealing with blackbody radiation problems, temperature scales as the fourth root of power, not linearly. The Stefan-Boltzmann relationship is key to connecting electrical power to radiated temperature.

Question 7

A furnace can be approximated as a blackbody cavity at 1500 K. A second identical furnace operates at 3000 K. If both have the same opening area, what is the ratio of the energy flux (power per unit area) emerging from the hot furnace to that from the cool furnace?

2
4
8
16 (correct answer)
32

Explanation: When you encounter blackbody radiation problems, you're dealing with the Stefan-Boltzmann law, which describes how much energy a perfect blackbody radiates. The key relationship is that energy flux (power per unit area) is proportional to the fourth power of absolute temperature:

j = \sigma T^4

, where

\sigma

is the Stefan-Boltzmann constant. Since both furnaces have identical opening areas and act as blackbody cavities, you can find the ratio by comparing their temperatures. The hot furnace operates at 3000 K and the cool furnace at 1500 K. The ratio of energy flux is:

\frac{j_{hot}}{j_{cool}} = \frac{\sigma T_{hot}^4}{\sigma T_{cool}^4} = \frac{T_{hot}^4}{T_{cool}^4} = \left(\frac{3000}{1500}\right)^4 = 2^4 = 16

Looking at the wrong answers: Choice A (2) represents just the temperature ratio, ignoring the fourth-power relationship entirely. Choice B (4) corresponds to a quadratic relationship (

T^2

), which might come from confusing this with other physics laws. Choice C (8) represents a cubic relationship (

T^3

), which has no basis in blackbody radiation theory. The correct answer is D (16) because blackbody energy flux scales with the fourth power of temperature. Study tip: Remember "T to the fourth" for blackbody radiation. This fourth-power dependence makes small temperature changes produce dramatic flux differences—a doubling of temperature increases energy output by a factor of 16, not 2.

Question 8

The intensity of blackbody radiation is maximum at a wavelength that depends on temperature according to Wien's displacement law. If a blackbody's temperature increases by 20%, by what percentage does the peak wavelength change?

Increases by 20%
Decreases by 16.7% (correct answer)
Decreases by 20%
Decreases by 44%
Increases by 44%

Explanation: Wien's displacement law describes how the peak wavelength of blackbody radiation shifts with temperature. This fundamental relationship appears frequently in thermodynamics and astrophysics problems, so understanding the inverse relationship between temperature and peak wavelength is crucial. Wien's displacement law states that

\lambda_{max} = \frac{b}{T}

, where

b

is Wien's constant and

T

is absolute temperature. This shows that peak wavelength is inversely proportional to temperature. If temperature increases by 20%, the new temperature becomes

T_{new} = 1.2T

. The new peak wavelength is:

\lambda_{new} = \frac{b}{1.2T} = \frac{\lambda_{original}}{1.2} = 0.833\lambda_{original}

This represents a decrease of

(1 - 0.833) \times 100\% = 16.7\%

, confirming answer B. Let's examine why the other options are incorrect: A) "Increases by 20%" ignores the inverse relationship entirely. Higher temperature means shorter wavelength, not longer. C) "Decreases by 20%" assumes a direct proportional relationship, as if wavelength decreases by the same percentage that temperature increases. This misses the inverse relationship. D) "Decreases by 44%" might result from incorrectly squaring the temperature change or making other mathematical errors with the inverse relationship. Remember: Wien's displacement law shows an inverse relationship between temperature and peak wavelength. When temperature increases, peak wavelength always decreases, but not by the same percentage due to this inverse proportionality. Always set up the fraction

\frac{1}{1 + \text{percentage change}}

when dealing with inverse relationships.

Question 9

A blackbody is heated from 1000 K to 2000 K. By what factor does the wavelength of peak emission change, and in which direction?

Increases by factor of 2
Increases by factor of 4
Remains the same
Decreases by factor of 2 (correct answer)
Decreases by factor of 4

Explanation: When you encounter blackbody radiation problems, you're dealing with Wien's displacement law, which describes how the peak emission wavelength changes with temperature. This law states that

\lambda_{max} = \frac{b}{T}

, where b is Wien's constant and T is absolute temperature. Since wavelength is inversely proportional to temperature, when temperature doubles from 1000 K to 2000 K, the peak wavelength must decrease. To find the factor, you can set up a ratio:

\frac{\lambda_2}{\lambda_1} = \frac{T_1}{T_2} = \frac{1000}{2000} = \frac{1}{2}

. This means the new wavelength is half the original, so it decreases by a factor of 2. Looking at the wrong answers: Choice A incorrectly suggests the wavelength increases by factor of 2, which would happen if wavelength were directly proportional to temperature—but it's inversely proportional. Choice B (increases by factor of 4) might come from confusing Wien's law with the Stefan-Boltzmann law, where total power radiated goes as

T^4

. Choice C (remains the same) ignores the temperature dependence entirely, which contradicts the fundamental physics of blackbody radiation. The correct answer is D: the wavelength decreases by a factor of 2. Study tip: Remember that Wien's displacement law shows an inverse relationship between temperature and peak wavelength. Hotter objects emit at shorter wavelengths (think: red-hot versus white-hot). This inverse relationship is key to solving these problems correctly.

Question 10

A red giant star has surface temperature 3000 K and a main sequence star has surface temperature 6000 K. Both stars radiate the same total power. What is the ratio of the red giant's radius to the main sequence star's radius?

0.25
0.5
2
4 (correct answer)
16

Explanation: When you encounter stellar physics problems involving temperature, radius, and power output, you're dealing with the Stefan-Boltzmann law, which relates a star's luminosity to its surface temperature and radius. The Stefan-Boltzmann law states that luminosity (total power) is

L = 4\pi R^2 \sigma T^4

, where R is radius, T is temperature, and σ is the Stefan-Boltzmann constant. Since both stars have the same luminosity, we can set up the equation:

L_{red} = L_{main}

4\pi R_{red}^2 \sigma T_{red}^4 = 4\pi R_{main}^2 \sigma T_{main}^4

Simplifying:

R_{red}^2 T_{red}^4 = R_{main}^2 T_{main}^4

Solving for the radius ratio:

\frac{R_{red}}{R_{main}} = \sqrt{\frac{T_{main}^4}{T_{red}^4}} = \frac{T_{main}^2}{T_{red}^2}

Substituting the temperatures:

\frac{R_{red}}{R_{main}} = \frac{(6000)^2}{(3000)^2} = \frac{36,000,000}{9,000,000} = 4

Answer D is correct. A) 0.25 represents the inverse ratio - this comes from incorrectly putting the red giant's temperature in the numerator. B) 0.5 results from using just the temperature ratio instead of squaring it. C) 2 comes from using the temperature ratio (6000/3000) without accounting for the T⁴ dependence in the Stefan-Boltzmann law. Remember: In stellar physics problems, temperature effects are amplified due to the T⁴ relationship. When comparing stars with the same luminosity, the cooler star must be significantly larger to compensate for its lower surface brightness.