Answers

ConceptChecks

ConceptCheck 5-1: The speed of light is incredibly fast, which made it very difficult to measure light’s speed except over enormous distances.

ConceptCheck 5-2: As Newton found when passing sunlight through a series of prisms, when one color is isolated from white light, there are no longer any other colors present in the remaining light. As a result, you cannot turn pure red into any other color. White light contains all of the colors, but once any of those colors are absorbed, they cannot be recovered. The red plastic absorbs all the visible wavelengths other than red, so no green can be obtained.

ConceptCheck 5-3: The width of your finger is about 1 cm, which falls in the range of the wavelength of microwaves (1 mm–10 cm).

ConceptCheck 5-4: Because the relationship between wavelength and frequency is c = λ × f, as one increases the other decreases (wavelength and frequency are inversely related). Thus, the longest wavelengths have the lowest frequencies and the shortest wavelengths have the highest frequencies.

ConceptCheck 5-5: As shown in Figure 5-12, at every wavelength, including infrared, the higher the temperature of a blackbody, the more energy it emits.

ConceptCheck 5-6: Yes. An object at 0°C still has a temperature of 237 K, and Kelvins are the temperature unit for working with blackbodies. At 273 K, the blackbody would peak in the infrared, although humans cannot see this light.

ConceptCheck 5-7: A star is hotter than our Sun if the star’s peak wavelength of blackbody emission is a shorter wavelength than the peak from our Sun.

ConceptCheck 5-8: Yes. If the cooler object is much larger than the warmer object, the cooler object can emit a greater total energy. The energy flux F refers to the energy emitted per second for each square meter of surface area, so even for a cool object, the larger the object, the more energy it radiates. This is why some stars (called red giants) that are much larger than our Sun are actually brighter than our Sun, even though they are cooler.

ConceptCheck 5-9: Photons with longer wavelengths will have lower energy than those with shorter wavelengths because the greater the wavelength, the lower the energy of a photon associated with that wavelength.

ConceptCheck 5-10: While still hot, the Sun’s outer atmosphere is cooler than deeper down where most of the Sun’s continuous blackbody emission originates. Thus, Figure 5-14 is an absorption line spectrum described by Kirchhoff’s third law. Kirchhoff’s second law would describe the emission line spectrum from hot helium gas, as in Figure 5-16.

ConceptCheck 5-11: An absorption spectra results when the light from a hot, dense object passes through the cooler, transparent gas of our atmosphere. At visible wavelengths, the atmosphere does not produce a significant absorption spectrum, but at infrared wavelengths (which are also emitted by hot lava), the atmosphere has many absorption lines.

ConceptCheck 5-12: The greater a photon’s energy, the higher its frequency. Since the change in energy for the n = 3 to n = 1 jump is the larger of the two transitions (see Figure 5-25), its emitted photon has the highest frequency. You can also compare the wavelengths of these two transitions in Figure 5-24, and the shorter the wavelength of light, the higher its frequency.

ConceptCheck 5-13: Yes. The most common visible-light hydrogen transition emits red photons of wavelength 656nm. It is an emission line (see Figure 5-24) with a transition from n = 3 to n = 2 and is referred to as H_α (H-alpha). This single transition accounts for most of the red seen in pretty nebulae such as Figure 5-19.

ConceptCheck 5-14: When the distance between an observer and a source is decreasing, the source’s entire spectrum will be shifted toward shorter wavelengths; alternatively, when the distance between an object and a source is increasing, the emissions lines will be shifted toward longer wavelengths. Thus, the star’s absorption lines will be observed at shorter wavelengths.

CalculationCheck

CalculationCheck 5-1: This classic rock station in Santa Barbara, California, emits waves with a frequency of 99.9 MHz, but this can be rounded to 100 MHz. To calculate the wavelength of these radio waves, we rearrange the equation c = ν ÷ λ to get λ = 3 × 10⁸ m/s ÷ 100 × 10⁶ Hz = 3.00 m.

CalculationCheck 5-2: Wien’s law can be rearranged to calculate the temperature of a star as T = 0.0029 K m ÷ (5800 K × 2) = 250 nm, which is ultraviolet.

CalculationCheck 5-3: ν = c × Δλ ÷ λ₀ = 3 × 10⁵ km/s × (486.3 nm − 486.2 nm) ÷ 486.2 nm = 61.7 km/s, and because it is moving toward longer wavelengths, the distance between the observer and the star must be increasing.