EM Radiation From Stars (OCR A Level Physics): Flashcards

Exam code: H556

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  • Define excitation.

Cards in this collection (55)

  • Define excitation.

    Excitation is the transition of an electron from a lower to a higher energy level, which requires energy to be absorbed.

  • Define de-excitation.

    De-excitation is the transition of an electron from a higher to a lower energy level, releasing energy as electromagnetic radiation of a specific frequency.

  • Give three ways energy can be provided for the excitation of a gas.

    • A photon of a specific frequency

    • Energy absorbed from the surroundings (heating)

    • Energy supplied by an electric field

  • Electrons in an atom can only occupy specific, .......... energy levels.

    Electrons in an atom can only occupy specific, discrete energy levels.

  • Why are electron energy levels given negative values?

    The energy of an electron is defined as zero when it is infinitely far from the nucleus. Since external energy is required to remove an electron from an energy level inside the atom, all energy values within the atom are negative, confining the electron to the atom.

  • Define ionisation.

    Ionisation is the complete removal of an electron from an atom.

  • What is the ground state, and what is its energy value for hydrogen?

    The ground state (n = 1) is the energy level with the most negative value, the lowest energy level available. For hydrogen, the ground state energy is –13.6 eV.

  • True or False?

    The energy value of an electron's energy level equals the energy released when the electron falls into that level.

    False.

    The value of an energy level equals the energy required to remove an electron from that level (its ionisation energy), not the energy released falling into it.

  • Define emission line spectrum.

    An emission line spectrum is produced when an excited electron moves from a higher to a lower energy level and emits a photon with energy equal to the difference between those energy levels.

  • Why does each element produce a unique emission line spectrum?

    Each element has a unique set of energy levels, so its electron transitions produce photons of specific, discrete wavelengths that act as a fingerprint of the element.

  • An emission line spectrum consists of a series of bright lines against a .......... background.

    An emission line spectrum consists of a series of bright lines against a dark background.

  • State the equation linking the energy of a photon to its frequency and to its wavelength.

    E = hf

    E = \frac{hc}{\lambda}

  • State the equation for the energy difference, ΔE, between two energy levels E1 and E2.

    \Delta E = E_{1} - E_{2} = hf = \frac{hc}{\lambda}

  • How does the wavelength of an emitted photon relate to the size of its energy level transition?

    Wavelength is inversely proportional to the size of the energy transition, so a larger energy transition produces a shorter wavelength.

  • True or False?

    In the hydrogen atom, transitions ending at n = 1 (the ground level) produce infrared radiation.

    False.

    Transitions ending at n = 1 produce ultraviolet radiation (highest energy, shortest wavelength). Transitions ending at n = 3 produce infrared radiation.

  • In hydrogen, what part of the electromagnetic spectrum do transitions ending at n = 2 produce?

    Visible light — violet is the highest energy transition in this series, and red is the lowest energy transition.

  • Why do stars produce absorption line spectra rather than emission line spectra?

    Photons produced in the star's core pass through cooler gas in the star's outer atmosphere. This atmosphere is not hot enough to produce an emission line spectrum, so an absorption line spectrum is seen instead.

  • Define absorption line spectrum.

    An absorption line spectrum is a continuous spectrum with dark lines, caused by a gas absorbing specific wavelengths of light as it passes through.

  • How can spectral lines in a star's absorption spectrum reveal its chemical composition?

    Each element produces a unique pattern of spectral lines (a fingerprint). If an element is present in the star, its characteristic pattern will appear as dark lines in the star's absorption spectrum.

  • Why do the missing wavelengths in an absorption spectrum appear as dark lines rather than being refilled by re-emitted photons?

    The re-emitted photons are emitted in all directions, rather than in the original direction of the white light, so those wavelengths appear to be missing from the observed spectrum.

  • The Sun is predominantly composed of .......... and helium gas.

    The Sun is predominantly composed of hydrogen and helium gas.

  • True or False?

    The dark lines in an element's absorption spectrum occur at different wavelengths from the bright lines in that element's emission spectrum.

    False.

    The wavelengths missing from an absorption spectrum are the same as those present in that element's emission line spectrum.

  • Define continuous spectrum.

    A continuous spectrum contains all wavelengths and frequencies of the electromagnetic spectrum. It is produced by hot, dense sources, such as the cores of stars.

  • What type of source produces an emission line spectrum, and why?

    A hot, low pressure gas. The atoms are far apart, so their energy levels are clearly quantised and well-defined, meaning only photons matching specific energy-level differences are emitted.

  • What conditions produce an absorption line spectrum?

    A hot, dense source (continuous spectrum) viewed through a cooler, low pressure gas, which absorbs light of specific wavelengths.

  • An absorption spectrum is represented by a continuous background with .......... lines at certain wavelengths.

    An absorption spectrum is represented by a continuous background with dark lines at certain wavelengths.

  • Why do hot, dense sources produce continuous spectra rather than line spectra?

    In a hot, dense material, atoms are so close together that they interact, spreading the energy states so they are not clearly defined. Photons of all frequencies are therefore emitted.

  • True or False?

    An emission line spectrum is represented by dark lines on a bright, coloured background.

    False.

    An emission line spectrum is represented by coloured lines on a black background — the opposite arrangement to an absorption spectrum.

  • Define transmission diffraction grating.

    A transmission diffraction grating is a glass or plastic slide containing a large number of regularly spaced, parallel slits or lines, used to analyse spectral line wavelengths from light emitted by stars.

  • Give two advantages of a diffraction grating over a prism or double slit for analysing starlight.

    • Greater angular dispersion (colour separation) than a prism

    • Produces sharper fringes than a double slit

  • What does N represent for a diffraction grating, and how is the slit spacing d found from it?

    N is the number of lines per unit length (e.g. per m). The slit spacing is found using d = \frac{1}{N}

  • Transmission diffraction gratings are useful for separating light of different wavelengths with high ...........

    Transmission diffraction gratings are useful for separating light of different wavelengths with high resolution.

  • State the diffraction grating equation.

    d\sin\theta = n\lambda

  • How is the highest order of maxima, n, determined for a diffraction grating?

    Using θ = 90o (so sin θ = 1) in the grating equation gives n = \frac{d}{\lambda}, rounded down to the nearest integer.

  • True or False?

    If the highest order calculated for a diffraction grating is n = 2.7, then three orders of maxima are visible.

    False.

    n must be an integer, and a decimal value is always rounded down, so only two orders (n = 2) are visible.

  • Define black body radiator.

    A black body radiator is an ideal object that absorbs and emits all wavelengths. Stars are the best real approximation to a black body.

  • State Wien's displacement law in words.

    The black body radiation curve for different temperatures peaks at a wavelength that is inversely proportional to the temperature.

  • According to Wien's law, the peak wavelength λmax is .......... proportional to the surface temperature T.

    According to Wien's law, the peak wavelength λmax is inversely proportional to the surface temperature T.

  • What does Wien's law imply about the colour of hotter and cooler stars?

    Hotter objects peak at a shorter wavelength, so appear white or blue. Cooler objects peak at a longer wavelength, so appear red or yellow.

  • What unit of temperature must be used in Wien's law calculations?

    Kelvin (K). Convert from degrees Celsius if the temperature is given in oC before using the equation.

  • True or False?

    A star with a shorter peak wavelength has a lower surface temperature than a star with a longer peak wavelength.

    False.

    Peak wavelength is inversely proportional to temperature, so a shorter peak wavelength corresponds to a higher surface temperature.

  • Define Stefan's law.

    Stefan's law (the Stefan-Boltzmann law) states that the total energy emitted by a black body per unit area per second is proportional to the fourth power of its absolute temperature.

  • State the equation for Stefan's law, defining each symbol.

    L = 4 \pi r^{2} \sigma T^{4}

    L = luminosity of the star (W)

    r = radius of the star

    σ = the Stefan-Boltzmann constant

    T = surface temperature of the star (K)

  • According to Stefan's law, how is a star's luminosity related to its radius and surface temperature?

    Luminosity is directly proportional to:

    • the square of the star's radius, L \propto r^{2}

    • the fourth power of the star's surface temperature, L \propto T^{4}

  • The total energy emitted by a black body per unit area per second is proportional to the .......... of the absolute temperature of the body.

    The total energy emitted by a black body per unit area per second is proportional to the fourth power of the absolute temperature of the body.

  • True or False?

    Luminosity is directly proportional to a star's surface temperature.

    False.

    Luminosity is proportional to the fourth power of the surface temperature, L \propto T^{4}, not to temperature itself.

  • What symbol is used for the Stefan-Boltzmann constant, and what is its approximate value?

    The Stefan-Boltzmann constant is represented by σ (sigma), with a value of 5.67 × 10-8 W m-2 K-4.

  • Star A and Star B have the same radius, but Star A has twice the surface temperature of Star B. According to Stefan's law, how does the luminosity of Star A compare to Star B?

    Star A is 16 times more luminous than Star B, since L \propto T^{4} and 2^{4} = 16.

  • What is the three-step procedure for estimating the radius of a star?

    1. Use Wien's displacement law to find the star's surface temperature

    2. Use the inverse square law of intensity to find the star's luminosity

    3. Use the Stefan-Boltzmann law to find the star's radius

  • Define luminosity of a star.

    Luminosity is the total power output of a star, equal to P in the intensity equation I = \frac{P}{A}.

  • State the inverse square law of intensity for a star, defining each symbol.

    I = \frac{L}{4 \pi d^{2}}

    I = intensity of light received on Earth (W m-2)

    L = luminosity of the star (W)

    d = distance between the star and Earth (m)

  • State Wien's displacement law for a star, defining each symbol.

    \lambda_{max} T = constant

    \lambda_{max} = wavelength emitted by the star at maximum intensity (m)

    T = surface temperature of the star (K)

  • The .......... law is used to calculate a star's luminosity from its intensity and its distance from Earth.

    The inverse square law is used to calculate a star's luminosity from its intensity and its distance from Earth.

  • True or False?

    In the equation I = \frac{L}{4 \pi d^{2}}, the area 4 \pi d^{2} is the surface area of the star.

    False.

    The area 4 \pi d^{2} is the surface area of the sphere over which the star's light has spread by the time it reaches Earth, not the star's own surface area.

  • A star has a peak wavelength half that of the Sun. According to Wien's displacement law, how does its surface temperature compare to the Sun's?

    The star's surface temperature is twice the Sun's, since \lambda_{max} T = constant means temperature is inversely proportional to peak wavelength.

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