Blackbody Radiation and the Quantum of Energy · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

Explaining a blackbody's spectrum

Science · Quantum & Modern Physics · ages 18-19
Name ______________________   Date ____________

  1. Heating a blackbody of fixed area increases its total emitted power and moves the peak of its power-per-unit-wavelength spectrum to shorter wavelengths.

    Circle one:   True   False

  2. Which statement defines an ideal blackbody?

    • It absorbs all incident radiation at every wavelength
    • It absorbs visible light but reflects every infrared wavelength
    • It emits no radiation even when its temperature is above zero kelvin

  3. What is the ultraviolet catastrophe in the classical Rayleigh-Jeans model?

    • A predicted absence of radiation throughout the ultraviolet band
    • A short-wavelength divergence giving infinite total emitted power
    • A predicted sharp ultraviolet line replacing the broad thermal spectrum

  4. A star is approximately a blackbody. Its power-per-unit-wavelength spectrum peaks at 560 nm. What can Wien's law estimate from that peak alone?

    • An estimate of its surface temperature
    • An estimate of its distance from Earth
    • An estimate of its age since formation

  5. Which change in Planck's oscillator model removes the short-wavelength divergence?

    • Every frequency receives equal energy regardless of its quantum size
    • Oscillator energy changes occur in discrete multiples of hf
    • All short-wavelength radiation is forbidden at every temperature

  6. How does a blackbody's total emitted power per unit area depend on its temperature T in kelvin?

    • It is proportional to the fourth power, T^4
    • It is proportional to the reciprocal temperature, 1/T
    • It is proportional to temperature measured in degrees Celsius

  7. Blackbody A's power-per-unit-wavelength spectrum peaks at 800 nm. Blackbody B's peaks at 400 nm. How do their kelvin temperatures compare?

    • A has twice B's temperature because its peak wavelength is twice as long
    • A and B have equal temperatures because both spectra contain many wavelengths
    • B has twice A's temperature because its peak wavelength is half as long

  8. A blackbody's power-per-unit-wavelength spectrum peaks at 966 nm. Use b = 2.898 x 10^-3 m K and Wien's law. What is its temperature in kelvin?

    Answer: ______________

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Answer key

For grown-ups. Fold this page away before handing over the rest.

Explaining a blackbody's spectrum W1-mt_B2Bk2IAOa9-s1

  1. True · Both changes occur: total emission increases, and the wavelength-spectrum peak moves to a shorter wavelength.
  2. It absorbs all incident radiation at every wavelength · A blackbody is a perfect absorber at every wavelength. It also emits thermal radiation according to its temperature.
  3. A short-wavelength divergence giving infinite total emitted power · The model grows without limit as wavelength approaches zero. Its total emitted power over the whole spectrum is infinite.
  4. An estimate of its surface temperature · Wien's law relates peak wavelength to temperature in kelvin. The peak alone does not give the star's age or distance.
  5. Oscillator energy changes occur in discrete multiples of hf · Discrete energy changes make high-frequency excitations less common at fixed temperature. They do not forbid short-wavelength radiation.
  6. It is proportional to the fourth power, T^4 · The Stefan-Boltzmann law gives P/A = sigma T^4. T must be measured in kelvin.
  7. B has twice A's temperature because its peak wavelength is half as long · Wien's law makes temperature inversely proportional to peak wavelength. Halving peak wavelength doubles kelvin temperature.
  8. 3000 · 966 nm = 9.66 x 10^-7 m. T = b / lambda_max = (2.898 x 10^-3) / (9.66 x 10^-7) = 3000 K.
Worksheet · LightMySky