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and its thermal emission produces the continuous spectrum observed from the star."},{"text":"The chromosphere (the thin, hot layer above the visible surface).","correct":false,"explanation":"The chromosphere lies above the photosphere and mainly contributes emission features. The continuous spectrum is produced by the dense, hot photosphere."},{"text":"The corona (the very hot outer atmosphere).","correct":false,"explanation":"The corona is a diffuse outer atmosphere that emits strongly at high-energy wavelengths, but its low density prevents it producing the star’s continuous visible spectrum."},{"text":"The core (the central region where fusion occurs).","correct":false,"explanation":"The core generates the star’s energy through nuclear fusion, but photons undergo many interactions before emerging. The continuous visible spectrum is produced at the photosphere."}]},{"id":"df8d0655-5e9c-4315-9bef-0eb435c8955d","cardType":"basic","points":3,"front":"Explain the formation of dark absorption lines in a stellar spectrum.","back":"

The photosphere emits a continuous spectrum
Photons are absorbed by atoms/ions in the cooler, tenuous atmosphere
This occurs at specific wavelengths corresponding to electron energy changes

","mcqOptions":[{"text":"

The photosphere emits a continuous spectrum
Photons are absorbed by atoms/ions in the cooler, tenuous atmosphere
This occurs at specific wavelengths corresponding to electron energy changes

","correct":true,"explanation":"A cooler, tenuous atmosphere absorbs photons at wavelengths matching allowed electron energy changes, removing those wavelengths from the continuous spectrum and producing dark lines."},{"text":"

The photosphere emits a line spectrum
Photons are emitted by atoms/ions in the cooler, tenuous atmosphere
This occurs at all wavelengths corresponding to electron energy changes

","correct":false,"explanation":"This reverses the process: the photosphere produces a continuous spectrum, while cooler atmospheric atoms absorb, rather than emit, photons at particular wavelengths."},{"text":"

The photosphere emits a continuous spectrum
Photons are scattered by atoms/ions in the cooler, tenuous atmosphere
This occurs at random wavelengths corresponding to electron energy changes

","correct":false,"explanation":"Scattering does not selectively remove photons through electron transitions. Absorption occurs at specific wavelengths determined by the atoms’ or ions’ allowed energy changes."},{"text":"

The photosphere emits a continuous spectrum
Photons are absorbed by atoms/ions in the hotter, dense atmosphere
This occurs at specific wavelengths corresponding to electron energy changes

","correct":false,"explanation":"The absorbing atmosphere must be cooler and relatively tenuous than the photosphere. Confusing it with a hotter, denser layer gives the wrong physical explanation for dark absorption lines."}]},{"id":"ae7cc14e-2b25-4c0f-bc0e-14381cd56f56","cardType":"cloze","points":null,"front":"According to $E = hf$, shorter wavelength photons (like X-rays) have [ ] energy than longer wavelength photons (like infrared).","back":"According to $E = hf$, shorter wavelength photons (like X-rays) have higher energy than longer wavelength photons (like infrared).","mcqOptions":[{"text":"higher","correct":true,"explanation":"Since $c = f\\lambda$, a shorter wavelength means a higher frequency. With $E = hf$, higher frequency photons have higher energy."},{"text":"lower","correct":false,"explanation":"This reverses the wavelength-frequency relationship. Shorter wavelengths have higher frequencies, so $E = hf$ shows that they have higher energy, not lower energy."},{"text":"the same","correct":false,"explanation":"Photon energy depends on frequency, which changes with wavelength. X-rays have higher frequencies than infrared photons and therefore do not have the same energy."},{"text":"zero","correct":false,"explanation":"A photon’s energy is not zero merely because its wavelength is short. Shorter wavelength means higher frequency and therefore greater energy through $E = hf$."}]},{"id":"a0de191f-da6f-4a4d-9ef4-f5f821721e5c","cardType":"basic","points":1,"front":"Define a black body.","back":"An ideal object that absorbs all incident electromagnetic radiation and emits a spectrum determined only by its temperature.","mcqOptions":[{"text":"An ideal object that absorbs all incident electromagnetic radiation and emits a spectrum determined only by its temperature.","correct":true,"explanation":"A black body is an ideal absorber that takes in all incident electromagnetic radiation. Its emitted spectrum depends only on its temperature, not its material or surface."},{"text":"An ideal object that reflects all incident electromagnetic radiation and emits a spectrum determined only by its temperature.","correct":false,"explanation":"A black body absorbs, rather than reflects, all incident electromagnetic radiation. Its emission is determined only by temperature, but complete reflection describes the opposite behaviour."},{"text":"An ideal object that absorbs all incident electromagnetic radiation and emits a spectrum determined only by its surface area.","correct":false,"explanation":"The defining emission property is dependence on temperature, not surface area. Surface area affects total power emitted, while temperature determines the spectrum’s shape."},{"text":"An ideal object that absorbs some incident electromagnetic radiation and emits a spectrum determined only by its composition.","correct":false,"explanation":"A black body absorbs all incident radiation and has an emission spectrum determined by temperature. Partial absorption and composition-dependent emission describe real materials rather than the ideal model."}]},{"id":"1f8ea77c-5e76-42eb-b9ab-be48c4a6664f","cardType":"basic","points":1,"front":"How do stars approximate black bodies despite having absorption lines?","back":"Their overall continuous spectrum closely matches the shape of a black-body curve.","mcqOptions":[{"text":"Their overall continuous spectrum closely matches the shape of a black-body curve.","correct":true,"explanation":"Absorption lines are narrow features superimposed on the broad continuum. When viewed overall, a star’s continuous spectrum has the characteristic shape of a black-body curve."},{"text":"Their overall line spectrum closely matches the shape of a black-body curve.","correct":false,"explanation":"The black-body comparison applies to the broad continuous background, not the individual absorption lines. The lines are superimposed on the overall continuous spectrum."},{"text":"Their overall continuous spectrum exactly matches the shape of a black-body curve.","correct":false,"explanation":"Stars approximate black bodies because their overall continuum closely matches the curve, but absorption lines and other atmospheric effects mean the match is not exact."},{"text":"Their individual absorption lines closely match the shape of a black-body curve.","correct":false,"explanation":"Individual absorption lines are narrow dips caused by cooler atmospheric atoms or ions. The broad underlying continuum, rather than each line, resembles a black-body curve."}]},{"id":"6c0791df-31b8-42af-8d2c-0fd1787127a0","cardType":"basic","points":1,"front":"State Wien's displacement law in words regarding temperature and wavelength.","back":"The peak wavelength ($\\lambda_{\\max}$) is inversely proportional to the absolute temperature.","mcqOptions":[{"text":"The peak wavelength ($\\lambda_{\\max}$) is inversely proportional to the absolute temperature.","correct":true,"explanation":"Wien’s displacement law gives $\\lambda_{\\max}T = \\text{constant}$. Increasing absolute temperature therefore shifts the peak to a shorter wavelength."},{"text":"The peak wavelength ($\\lambda_{\\max}$) is directly proportional to the absolute temperature.","correct":false,"explanation":"This reverses Wien’s law. Because $\\lambda_{\\max}T$ is constant, a higher absolute temperature produces a shorter, not longer, peak wavelength."},{"text":"The peak frequency ($f_{\\max}$) is inversely proportional to the absolute temperature.","correct":false,"explanation":"Wien’s displacement law is stated for peak wavelength, not peak frequency. The relevant relationship is between $\\lambda_{\\max}$ and absolute temperature."},{"text":"The peak wavelength ($\\lambda_{\\max}$) is inversely proportional to the Celsius temperature.","correct":false,"explanation":"The temperature must be measured on an absolute scale, such as kelvin. Celsius temperatures do not give the proportional relationship required by Wien’s displacement law."}]},{"id":"e296940e-ba50-4be6-b99c-7ca2f0a312ff","cardType":"cloze","points":null,"front":"The constant for Wien's displacement law is $\\lambda_{\\max}T = $ $\\text{[...]}$ m K.","back":"The constant for Wien's displacement law is $\\lambda_{\\max}T = $ $\\boxed{2.90 \\times 10^{-3}}$ m K.","mcqOptions":[{"text":"$$2.90 \\times 10^{-3}$","correct":true,"explanation":"Wien’s displacement constant is approximately $2.90 \\times 10^{-3}\\ \\text{m K}$, so multiplying the peak wavelength in metres by temperature in kelvin gives this value."},{"text":"$$2.90 \\times 10^{3}$","correct":false,"explanation":"The exponent has the wrong sign, making the value enormously too large. The displacement constant is approximately $2.90 \\times 10^{-3}\\ \\text{m K}$."},{"text":"$$2.90 \\times 10^{-6}$","correct":false,"explanation":"This uses an exponent three powers too small, often from confusing the standard constant with a micrometre conversion. The correct Wien constant is $2.90 \\times 10^{-3}\\ \\text{m K}$."},{"text":"$$3.00 \\times 10^{-3}$","correct":false,"explanation":"This is a tempting rounded approximation, but the stated value for Wien’s displacement constant is $2.90 \\times 10^{-3}\\ \\text{m K}$, not $3.00 \\times 10^{-3}$."}]},{"id":"7566423c-3e15-44c9-84ac-50f5c17dcbf1","cardType":"basic","points":2,"front":"Contrast luminosity and flux.","back":"

Luminosity ($L$): Total power emitted by the star (Watts)
Flux ($F$): Power received per unit area at a distance (W m^-2)

","mcqOptions":[{"text":"

Luminosity ($L$): Total power emitted by the star (Watts)
Flux ($F$): Power received per unit area at a distance (W m^-2)

","correct":true,"explanation":"Luminosity is the star’s total emitted power, measured in watts. Flux is the power received per unit area and decreases with distance as the radiation spreads out."},{"text":"

Luminosity ($L$): Power received per unit area at a distance (W m^-2)
Flux ($F$): Total power emitted by the star (Watts)

","correct":false,"explanation":"These definitions are reversed. Luminosity is total emitted power in watts, whereas flux is received power per unit area in W m^-2."},{"text":"

Luminosity ($L$): Total power emitted by the star (W m^-2)
Flux ($F$): Power received per unit area at a distance (Watts)

","correct":false,"explanation":"The physical descriptions are correct, but the units are swapped. Total power has units of watts, while power per unit area has units of W m^-2."},{"text":"

Luminosity ($L$): Total energy emitted by the star (Joules)
Flux ($F$): Total power received at a distance (Watts)

","correct":false,"explanation":"Luminosity is a rate of energy transfer, so it is total power in watts, and flux must be per unit area in W m^-2. Energy alone and total received power are different quantities."}]},{"id":"85a0b377-3c5f-43b3-a308-c15d60e97b06","cardType":"cloze","points":null,"front":"In radiation laws like Stefan's and Wien's, temperature must always be measured in [ ].","back":"In radiation laws like Stefan's and Wien's, temperature must always be measured in kelvin (K).","mcqOptions":[{"text":"kelvin (K)","correct":true,"explanation":"Kelvin is the absolute temperature scale required by Stefan’s and Wien’s laws, because these relationships use temperature measured from absolute zero rather than degrees Celsius."},{"text":"degrees Celsius (°C)","correct":false,"explanation":"Celsius temperatures cannot be substituted directly into radiation laws because the scale does not start at absolute zero. The correct scale is kelvin (K)."},{"text":"degrees Fahrenheit (°F)","correct":false,"explanation":"Fahrenheit is not an absolute temperature scale and is therefore unsuitable for these laws. Temperature must be converted to kelvin (K) before calculation."},{"text":"joules (J)","correct":false,"explanation":"Radiation laws require an absolute temperature, not a temperature expressed as an energy unit. The required unit is kelvin (K)."}]},{"id":"60b8f9cd-2bcd-4b60-ae46-cf40e9a964ef","cardType":"basic","points":1,"front":"State Stefan's law for a spherical star of radius $R$.","back":"$$L = 4\\pi R^2 \\sigma T^4$","mcqOptions":[{"text":"$$L = 4\\pi R^2 \\sigma T^4$","correct":true,"explanation":"A spherical star emits from surface area $4\\pi R^2$, and Stefan’s law gives emitted power per unit area as $\\sigma T^4$, so multiplying them gives the luminosity."},{"text":"$$L = 4R^2 \\sigma T^4$","correct":false,"explanation":"This omits the factor $\\pi$, which is needed when the spherical surface area and Stefan’s law are combined. The correct relationship is $L = 4\\pi R^2 \\sigma T^4$."},{"text":"$$L = 4\\pi R \\sigma T^4$","correct":false,"explanation":"Using $4\\pi R$ confuses a length with the spherical surface area, which depends on the radius squared. The correct relationship is $L = 4\\pi R^2 \\sigma T^4$."},{"text":"$$L = 4\\pi R^2 \\sigma T^2$","correct":false,"explanation":"The temperature dependence in Stefan’s law is fourth power, not squared. The correct relationship is $L = 4\\pi R^2 \\sigma T^4$."}]},{"id":"5875339d-47b2-4753-a96c-6c792a438ea4","cardType":"basic","points":2,"front":"If a star's absolute temperature doubles, by what factor does its luminosity increase (assuming $R$ is constant)?","back":"16 (calculated as $2^4$).","mcqOptions":[{"text":"16 (calculated as $2^4$).","correct":true,"explanation":"Stefan’s law gives $L \\propto T^4$ when radius is constant. Doubling temperature therefore multiplies luminosity by $2^4 = 16$."},{"text":"2 (calculated as $2^1$).","correct":false,"explanation":"A factor of 2 would result from a direct proportionality, but luminosity depends on the fourth power of temperature. Doubling temperature gives $2^4 = 16$."},{"text":"4 (calculated as $2^2$).","correct":false,"explanation":"A factor of 4 uses a square-law dependence, which is not Stefan’s temperature dependence. Since $L \\propto T^4$, doubling temperature gives 16."},{"text":"8 (calculated as $2^3$).","correct":false,"explanation":"A factor of 8 uses a cube-law dependence and ignores the fourth power in Stefan’s law. Doubling temperature gives $2^4 = 16$."}]},{"id":"c03217d2-5353-423d-83c7-85b1598602a6","cardType":"basic","points":1,"front":"State the inverse square law for radiation flux.","back":"$$F = \\frac{L}{4\\pi d^2}$","mcqOptions":[{"text":"$$F = \\frac{L}{4\\pi d^2}$","correct":true,"explanation":"The flux decreases with the square of distance because the same luminosity spreads over a spherical area $4\\pi d^2$. Therefore $F = \\frac{L}{4\\pi d^2}$."},{"text":"$$F = \\frac{L}{4\\pi d}$","correct":false,"explanation":"This expression uses distance rather than distance squared, giving a linear decrease instead of the inverse square decrease. The correct law is $F = \\frac{L}{4\\pi d^2}$."},{"text":"$$F = \\frac{L d^2}{4\\pi}$","correct":false,"explanation":"Multiplying by distance squared predicts that flux increases farther from the source, reversing the physical relationship. The correct law divides by $4\\pi d^2$."},{"text":"$$F = \\frac{4\\pi L}{d^2}$","correct":false,"explanation":"The factor $4\\pi$ belongs in the denominator because radiation spreads over a sphere. The correct expression is $F = \\frac{L}{4\\pi d^2}$."}]},{"id":"a90c2b3d-994e-41a3-92fe-b90c907068be","cardType":"basic","points":3,"front":"Outline the steps to determine a star's physical radius ($R$) from Earth-based observations.","back":"

Find $T$ using Wien's law and peak wavelength
Find $L$ using the inverse square law with flux and distance
Calculate $R$ using Stefan's law ($L = 4\\pi R^2 \\sigma T^4$)

","mcqOptions":[{"text":"

Find $T$ using Wien's law and peak wavelength
Find $L$ using the inverse square law with flux and distance
Calculate $R$ using Stefan's law ($L = 4\\pi R^2 \\sigma T^4$)

","correct":true,"explanation":"Wien’s law and the peak wavelength give $T$; flux, distance, and the inverse square law give $L$; Stefan’s law then gives the physical radius $R$."},{"text":"

Find $T$ using the inverse square law and distance
Find $L$ using Wien's law and peak wavelength
Calculate $R$ using the Stefan's law ($L = 4\\pi R^2 \\sigma T^4$)

","correct":false,"explanation":"This reverses the first two observations: temperature comes from the peak wavelength, while luminosity comes from flux and distance. Stefan’s law is then used to calculate $R$."},{"text":"

Find $R$ using Wien's law and peak wavelength
Find $L$ using the inverse square law with flux and distance
Calculate $T$ using Stefan's law ($L = 4\\pi R^2 \\sigma T^4$)

","correct":false,"explanation":"Using Wien’s law to find radius is a category error: it determines temperature from wavelength. Radius is obtained only after finding luminosity and temperature with Stefan’s law."},{"text":"

Find $T$ using Stefan's law and peak wavelength
Find $R$ using the inverse square law with flux and distance
Calculate $L$ using Wien's law ($L = 4\\pi R^2 \\sigma T^4$)

","correct":false,"explanation":"The inverse square law determines luminosity from observed flux and distance, not radius. Wien’s law finds temperature, after which Stefan’s law can determine $R$."}]},{"id":"6fe31c30-a337-4517-8515-09e5064aa025","cardType":"cloze","points":null,"front":"To find the Stefan constant $\\sigma$ from a graph, plot emitted power per unit area against $\\text{[...]}$.","back":"To find the Stefan constant $\\sigma$ from a graph, plot emitted power per unit area against $\\boxed{T^4}$.","mcqOptions":[{"text":"$$T^4$","correct":true,"explanation":"Stefan’s law states that emitted power per unit area is proportional to $T^4$. A graph against $T^4$ therefore has gradient equal to the Stefan constant $\\sigma$."},{"text":"$$T$","correct":false,"explanation":"Plotting against $T$ would not produce the linear relationship required by Stefan’s law, because emitted power per unit area is proportional to the fourth power of temperature. The correct quantity is $T^4$."},{"text":"$$T^2$","correct":false,"explanation":"The square is too low a power: Stefan’s law uses a fourth-power dependence. Plotting against $T^2$ would not give the correct gradient for $\\sigma$."},{"text":"$$\\frac{1}{T^4}$","correct":false,"explanation":"The reciprocal gives the wrong relationship and would make the graph unsuitable for obtaining the Stefan constant as its gradient. The correct horizontal-axis quantity is $T^4$."}]},{"id":"98952b95-966a-4639-a457-7a41402178bd","cardType":"basic","points":2,"front":"Why is multiwavelength astronomy necessary to understand star formation in dusty clouds?","back":"Dust absorbs visible light, but longer infrared wavelengths can pass through to reveal cool protostars within.","mcqOptions":[{"text":"Dust absorbs visible light, but longer infrared wavelengths can pass through to reveal cool protostars within.","correct":true,"explanation":"Dust absorbs visible light, hiding young stars and protostars. Infrared radiation has longer wavelengths that can pass through the dust and reveal these cool objects."},{"text":"Dust absorbs infrared, but shorter visible wavelengths can pass through to reveal cool protostars within.","correct":false,"explanation":"Dust does not mainly block infrared while transmitting visible light; it is visible light that is strongly absorbed. Infrared observations can reveal cool protostars through the dust."},{"text":"Dust absorbs ultraviolet light, but longer radio wavelengths can pass through to reveal cool protostars within.","correct":false,"explanation":"Ultraviolet radiation has shorter wavelengths than visible light and is also absorbed or scattered by dust. Infrared wavelengths are used to observe the hidden protostars."},{"text":"Dust absorbs visible light, but shorter ultraviolet wavelengths can pass through to reveal cool protostars within.","correct":false,"explanation":"Dusty clouds are not made transparent by using shorter wavelengths. Visible light is absorbed, whereas longer infrared wavelengths can pass through and reveal protostars."}]},{"id":"936e3524-3e1b-4263-b551-252082034415","cardType":"cloze","points":null,"front":"In multiwavelength astronomy, radio waves are used to map [ ].","back":"In multiwavelength astronomy, radio waves are used to map cold gas clouds.","mcqOptions":[{"text":"cold gas clouds","correct":true,"explanation":"Cold gas clouds emit and interact strongly with radio wavelengths, allowing astronomers to map their distribution even when visible light is obscured by dust."},{"text":"hot stellar surfaces","correct":false,"explanation":"Radio observations can map gas, but the hottest stars are more effectively studied at shorter wavelengths. The specific target here is cold gas clouds."},{"text":"warm dust grains","correct":false,"explanation":"Infrared wavelengths are particularly useful for detecting cool protostars and warm dust, not for the stated radio mapping target. Radio waves map cold gas clouds."},{"text":"the cosmic microwave background","correct":false,"explanation":"The cosmic microwave background is observed at microwave wavelengths, but this is not the main object mapped by radio observations in star-forming regions. The correct term is cold gas clouds."}]},{"id":"b119cd2a-a093-4bc3-b8e7-d30a80d3084c","cardType":"basic","points":1,"front":"What physical processes are typically revealed by X-ray and gamma ray observations?","back":"Very energetic processes, such as supernova remnants or material near compact objects.","mcqOptions":[{"text":"Very energetic processes, such as supernova remnants or material near compact objects.","correct":true,"explanation":"X-rays and gamma rays are produced by extremely hot, rapidly moving or strongly accelerated matter, so they trace supernova remnants, accretion near compact objects and similarly energetic events."},{"text":"Cool photospheric processes, such as molecular absorption or convection in ordinary stars.","correct":false,"explanation":"Cool photospheric convection and molecular absorption mainly affect infrared or visible radiation. X-ray and gamma ray observations instead reveal energetic processes such as supernova remnants or matter near compact objects."},{"text":"Low-energy atomic transitions, such as visible absorption by neutral atoms in stellar atmospheres.","correct":false,"explanation":"Low-energy transitions in neutral atoms generally produce visible or infrared spectral features. X-rays and gamma rays instead trace highly energetic phenomena, including supernova remnants and accretion near compact objects."},{"text":"Long-period orbital processes, such as planetary transits or eclipses in binary systems.","correct":false,"explanation":"Planetary transits and eclipses are detected through periodic changes in visible brightness, not primarily through high-energy radiation. X-rays and gamma rays reveal energetic processes near extreme astronomical objects."}]},{"id":"669c84f0-2a69-441b-a0db-a7cdb5609eb6","cardType":"cloze","points":null,"front":"On a graph testing the inverse square law, flux ($F$) plotted against $\\text{[...]}$ should yield a straight line through the origin.","back":"On a graph testing the inverse square law, flux ($F$) plotted against $\\boxed{1/d^2}$ should yield a straight line through the origin.","mcqOptions":[{"text":"$$1/d^2$","correct":true,"explanation":"For a source of constant luminosity, $F = L/(4\\pi d^2)$, so flux is directly proportional to $1/d^2$. The resulting graph is therefore linear through the origin."},{"text":"$$d^2$","correct":false,"explanation":"This reverses the inverse-square relationship and would imply that flux increases with distance. Since $F \\propto 1/d^2$, the horizontal axis must be $1/d^2$."},{"text":"$$1/d$","correct":false,"explanation":"This treats flux as following a simple inverse relationship and misses the geometrical spreading over an area proportional to $d^2$. A linear plot requires $1/d^2$."},{"text":"$$d$","correct":false,"explanation":"This assumes flux is directly proportional to distance, whereas radiation spreads over an increasing spherical area. Because $F \\propto 1/d^2$, flux should be plotted against $1/d^2$."}]},{"id":"a5b3f5b1-4a31-4445-9366-33fef20e227a","cardType":"basic","points":1,"front":"How does the uncertainty in peak wavelength ($\\lambda_{\\max}$) relate to the uncertainty in calculated temperature ($T$)?","back":"The percentage uncertainty in $T$ is approximately equal to the percentage uncertainty in $\\lambda_{\\max}$.","mcqOptions":[{"text":"The percentage uncertainty in $T$ is approximately equal to the percentage uncertainty in $\\lambda_{\\max}$.","correct":true,"explanation":"Wien’s law gives $T = b/\\lambda_{\\max}$. For a reciprocal relationship, the fractional or percentage uncertainties have approximately equal magnitudes when the uncertainty in $b$ is negligible."},{"text":"The absolute uncertainty in $T$ is approximately equal to the absolute uncertainty in $\\lambda_{\\max}$.","correct":false,"explanation":"Absolute uncertainties cannot be equated because temperature and wavelength have different units and scales. From $T = b/\\lambda_{\\max}$, their percentage uncertainties are approximately equal."},{"text":"The percentage uncertainty in $T$ is approximately twice the percentage uncertainty in $\\lambda_{\\max}$.","correct":false,"explanation":"A factor of two would arise from an inverse-square dependence, but Wien’s law contains the first power of wavelength. Therefore, the percentage uncertainties in $T$ and $\\lambda_{\\max}$ are approximately equal."},{"text":"The percentage uncertainty in $T$ is approximately half the percentage uncertainty in $\\lambda_{\\max}$.","correct":false,"explanation":"Halving the uncertainty incorrectly treats the reciprocal relationship like a square-root dependence. Since $T \\propto \\lambda_{\\max}^{-1}$, the two percentage uncertainties have approximately equal magnitudes."}]},{"id":"27111dd3-9e4e-4af8-8271-606ab3499347","cardType":"basic","points":1,"front":"What is the primary way astronomers identify specific elements in a star's atmosphere?","back":"By matching the pattern of absorption lines to laboratory spectra of known elements.","mcqOptions":[{"text":"By matching the pattern of absorption lines to laboratory spectra of known elements.","correct":true,"explanation":"Each element has a distinctive set of allowed electron transitions and therefore a characteristic absorption-line pattern. Comparing stellar lines with laboratory spectra reveals which elements are present in the atmosphere."},{"text":"By fitting the shape of the continuous spectrum to black-body curves at known temperatures.","correct":false,"explanation":"The continuous spectrum mainly indicates the star’s surface temperature through its black-body shape. Elemental composition is instead identified by matching absorption-line patterns with laboratory spectra of known elements."},{"text":"By measuring the displacement of spectral lines to calculate the star's radial velocity.","correct":false,"explanation":"Displacement of spectral lines from their laboratory wavelengths reveals radial velocity through the Doppler effect. The elements themselves are identified from the pattern of absorption lines, not their overall displacement."},{"text":"By comparing the star's apparent brightness with standard candles of known luminosity.","correct":false,"explanation":"Comparing apparent brightness with a known luminosity can provide information about distance, not atmospheric composition. Specific elements are identified by matching their absorption-line patterns to laboratory spectra."}]}]}],"examCountdown":{"examDate":"2027-05-21","paperLabel":"Component 1"},"heading":"Using Radiation to Investigate Stars","title":"Using Radiation to Investigate Stars Flashcards","topbar":{"backHref":"/a-level/physics/eduqas/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"0ccc76c3-6fb4-547b-ac75-f5f7809142e0","walled":"$L39"}],"$L3a","$L3b"]