Astronomy Quiz: Electromagnetic Spectrum
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Electromagnetic SpectrumQuestion 1 of 20

An instrument detects four photons: Photon W with a frequency of 2.0×10152.0 \times 10^{15} Hz, Photon X with a wavelength of 650 nm, Photon Y with an energy of 2.5 eV, and Photon Z which is known to be a microwave photon. Which of the following lists the photons in order from highest energy to lowest energy?

W, Y, X, Z
Z, X, Y, W
W, X, Y, Z
Y, W, X, Z
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Astronomy Quiz

Astronomy Quiz: Electromagnetic Spectrum

Practice Electromagnetic Spectrum in Astronomy 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 Electromagnetic Spectrum, giving you a quick way to practice the rules, question types, and explanations that matter most for Astronomy.

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.

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Question 1

An instrument detects four photons: Photon W with a frequency of 2.0×10152.0 \times 10^{15} Hz, Photon X with a wavelength of 650 nm, Photon Y with an energy of 2.5 eV, and Photon Z which is known to be a microwave photon. Which of the following lists the photons in order from highest energy to lowest energy?

  1. W, Y, X, Z (correct answer)
  2. Z, X, Y, W
  3. W, X, Y, Z
  4. Y, W, X, Z
Explanation: To compare the photons, we should convert their properties to a common unit, such as electron volts (eV). Use E=hνE = h\nu and E=hc/λE = hc/\lambda, with h≈4.14×10−15h \approx 4.14 \times 10^{-15} eV·s and hc≈1240hc \approx 1240 eV·nm.
  • Photon W: E=(4.14×10−15 eV⋅s)(2.0×1015 Hz)=8.28E = (4.14 \times 10^{-15} \text{ eV}\cdot\text{s})(2.0 \times 10^{15} \text{ Hz}) = 8.28 eV (Ultraviolet).
  • Photon X: E=1240 eV⋅nm/650 nm≈1.91E = 1240 \text{ eV}\cdot\text{nm} / 650 \text{ nm} \approx 1.91 eV (Red light).
  • Photon Y: E=2.5E = 2.5 eV (Green light).
  • Photon Z: Microwaves have much longer wavelengths and therefore much lower energies than visible or UV light (typically 10−610^{-6} to 10−310^{-3} eV). The order from highest to lowest energy is W (8.28 eV) > Y (2.5 eV) > X (1.91 eV) > Z (< 1 eV).

Question 2

A space telescope uses a filter that transmits photons with energies between 2.2 eV and 2.8 eV. An astronomer uses this telescope to observe a star. Which colors of light from the star will be most prominently detected? (Note: The product of Planck's constant and the speed of light, hchc, is approximately 1240 eV·nm).

  1. A band in the red and orange part of the spectrum.
  2. A band in the blue and green part of the spectrum. (correct answer)
  3. Only ultraviolet light, as these energies are too high for visible light.
  4. The entire visible spectrum, from violet to red.
Explanation: First, convert the energy range to a wavelength range using the formula λ=hc/E\lambda = hc/E. The minimum wavelength is λmin=1240 eV⋅nm/2.8 eV≈443\lambda_{min} = 1240 \text{ eV}\cdot\text{nm} / 2.8 \text{ eV} \approx 443 nm. The maximum wavelength is λmax=1240 eV⋅nm/2.2 eV≈564\lambda_{max} = 1240 \text{ eV}\cdot\text{nm} / 2.2 \text{ eV} \approx 564 nm. This wavelength range of approximately 443 nm to 564 nm corresponds to the violet, blue, and green parts of the visible spectrum. Among the choices, 'blue and green' is the best description.

Question 3

An instrument detects four photons: Photon W with a frequency of 2.0×10152.0 \times 10^{15} Hz, Photon X with a wavelength of 650 nm, Photon Y with an energy of 2.5 eV, and Photon Z which is known to be a microwave photon. Which of the following lists the photons in order from highest energy to lowest energy?

  1. W, Y, X, Z (correct answer)
  2. Z, X, Y, W
  3. W, X, Y, Z
  4. Y, W, X, Z
Explanation: To compare the photons, we should convert their properties to a common unit, such as electron volts (eV). Use E=hνE = h\nu and E=hc/λE = hc/\lambda, with h≈4.14×10−15h \approx 4.14 \times 10^{-15} eV·s and hc≈1240hc \approx 1240 eV·nm.
  • Photon W: E=(4.14×10−15 eV⋅s)(2.0×1015 Hz)=8.28E = (4.14 \times 10^{-15} \text{ eV}\cdot\text{s})(2.0 \times 10^{15} \text{ Hz}) = 8.28 eV (Ultraviolet).
  • Photon X: E=1240 eV⋅nm/650 nm≈1.91E = 1240 \text{ eV}\cdot\text{nm} / 650 \text{ nm} \approx 1.91 eV (Red light).
  • Photon Y: E=2.5E = 2.5 eV (Green light).
  • Photon Z: Microwaves have much longer wavelengths and therefore much lower energies than visible or UV light (typically 10−610^{-6} to 10−310^{-3} eV). The order from highest to lowest energy is W (8.28 eV) > Y (2.5 eV) > X (1.91 eV) > Z (< 1 eV).

Question 4

The energy required to ionize a hydrogen atom from its ground state is 13.6 eV. What is the maximum wavelength a photon can have and still be capable of causing this ionization, and in what region of the spectrum does this wavelength lie? (Note: hc≈1240hc \approx 1240 eV·nm).

  1. 91 nm; Ultraviolet (correct answer)
  2. 16864 nm; Infrared
  3. 91 nm; X-ray
  4. 1240 nm; Infrared
Explanation: To ionize the atom, a photon must have a minimum energy of 13.6 eV. The maximum wavelength corresponds to this minimum energy. Using the relation λ=hc/E\lambda = hc/E, we find λmax=1240 eV⋅nm/13.6 eV≈91.2\lambda_{max} = 1240 \text{ eV}\cdot\text{nm} / 13.6 \text{ eV} \approx 91.2 nm. Wavelengths shorter than this will also have enough energy. The electromagnetic spectrum places wavelengths less than 400 nm but greater than 10 nm in the ultraviolet (UV) region. Therefore, the ionization threshold for hydrogen is at 91 nm, in the UV.

Question 5

The energy required to ionize a hydrogen atom from its ground state is 13.6 eV. What is the maximum wavelength a photon can have and still be capable of causing this ionization, and in what region of the spectrum does this wavelength lie? (Note: hc≈1240hc \approx 1240 eV·nm).

  1. 91 nm; Ultraviolet (correct answer)
  2. 16864 nm; Infrared
  3. 91 nm; X-ray
  4. 1240 nm; Infrared
Explanation: To ionize the atom, a photon must have a minimum energy of 13.6 eV. The maximum wavelength corresponds to this minimum energy. Using the relation λ=hc/E\lambda = hc/E, we find λmax=1240 eV⋅nm/13.6 eV≈91.2\lambda_{max} = 1240 \text{ eV}\cdot\text{nm} / 13.6 \text{ eV} \approx 91.2 nm. Wavelengths shorter than this will also have enough energy. The electromagnetic spectrum places wavelengths less than 400 nm but greater than 10 nm in the ultraviolet (UV) region. Therefore, the ionization threshold for hydrogen is at 91 nm, in the UV.

Question 6

In a hydrogen atom, an electron transition from n=3 to n=2 results in the emission of a photon (H-alpha). A different transition from n=2 to n=1 results in the emission of another photon (Lyman-alpha). Which statement correctly compares the wavelengths of these two photons?

  1. The H-alpha photon has a shorter wavelength because its transition originates from a higher energy level (n=3).
  2. The Lyman-alpha photon has a shorter wavelength because the energy difference between n=2 and n=1 is greater than between n=3 and n=2. (correct answer)
  3. The photons have equal wavelengths because both transitions are between adjacent energy levels.
  4. The Lyman-alpha photon has a longer wavelength because its transition ends in the ground state (n=1).
Explanation: The energy of an emitted photon is equal to the energy difference between the initial and final atomic energy levels. The energy levels in a hydrogen atom are given by En∝−1/n2E_n \propto -1/n^2, which means the energy gaps are larger at lower n values. The energy drop from n=2 to n=1 is significantly larger than the energy drop from n=3 to n=2. Since photon energy is inversely proportional to wavelength (E=hc/λE=hc/\lambda), the higher-energy Lyman-alpha photon will have a shorter wavelength than the lower-energy H-alpha photon.

Question 7

A beam of red light (λ=700\lambda = 700 nm) and a beam of blue light (λ=400\lambda = 400 nm) are measured by a detector. The total power (energy per second) of the red beam is measured to be equal to the total power of the blue beam. If the red beam consists of NRN_R photons per second and the blue beam consists of NBN_B photons per second, what is the ratio NR/NBN_R / N_B?

  1. 4/7 (approximately 0.57)
  2. 1
  3. 7/4 (or 1.75) (correct answer)
  4. 49/16 (approximately 3.06)
Explanation: The total power (P) of a beam of light is the number of photons per second (N) multiplied by the energy of a single photon (E). So, P=N×E=N×(hc/λ)P = N \times E = N \times (hc/\lambda). We are given that Pred=PblueP_{red} = P_{blue}. Therefore, NR×(hc/λR)=NB×(hc/λB)N_R \times (hc/\lambda_R) = N_B \times (hc/\lambda_B). The hchc term cancels out, leaving NR/λR=NB/λBN_R/\lambda_R = N_B/\lambda_B. To find the ratio NR/NBN_R / N_B, we rearrange the equation: NR/NB=λR/λBN_R / N_B = \lambda_R / \lambda_B. Plugging in the values gives 700 nm/400 nm=7/4=1.75700 \text{ nm} / 400 \text{ nm} = 7/4 = 1.75. This means that 1.75 times more red photons are needed to deliver the same total energy as the more energetic blue photons.

Question 8

A beam of red light (λ=700\lambda = 700 nm) and a beam of blue light (λ=400\lambda = 400 nm) are measured by a detector. The total power (energy per second) of the red beam is measured to be equal to the total power of the blue beam. If the red beam consists of NRN_R photons per second and the blue beam consists of NBN_B photons per second, what is the ratio NR/NBN_R / N_B?

  1. 4/7 (approximately 0.57)
  2. 1
  3. 7/4 (or 1.75) (correct answer)
  4. 49/16 (approximately 3.06)
Explanation: The total power (P) of a beam of light is the number of photons per second (N) multiplied by the energy of a single photon (E). So, P=N×E=N×(hc/λ)P = N \times E = N \times (hc/\lambda). We are given that Pred=PblueP_{red} = P_{blue}. Therefore, NR×(hc/λR)=NB×(hc/λB)N_R \times (hc/\lambda_R) = N_B \times (hc/\lambda_B). The hchc term cancels out, leaving NR/λR=NB/λBN_R/\lambda_R = N_B/\lambda_B. To find the ratio NR/NBN_R / N_B, we rearrange the equation: NR/NB=λR/λBN_R / N_B = \lambda_R / \lambda_B. Plugging in the values gives 700 nm/400 nm=7/4=1.75700 \text{ nm} / 400 \text{ nm} = 7/4 = 1.75. This means that 1.75 times more red photons are needed to deliver the same total energy as the more energetic blue photons.

Question 9

A radio telescope observes a signal from neutral hydrogen gas at a frequency of 1.42 GHz (1.42×1091.42 \times 10^9 Hz). A different space telescope observes an X-ray source emitting photons with an energy that is 10910^9 times greater than the energy of the radio photons. What is the approximate wavelength of these X-ray photons? (Speed of light c≈3×108c \approx 3 \times 10^8 m/s)

  1. 2.1×10−102.1 \times 10^{-10} m (correct answer)
  2. 4.7×10−184.7 \times 10^{-18} m
  3. 2.1×1082.1 \times 10^8 m
  4. 4.7×1004.7 \times 10^0 m
Explanation: This is a two-step problem. First, find the wavelength of the radio photon using λ=c/ν\lambda = c/\nu. λradio=(3×108 m/s)/(1.42×109 Hz)≈0.21\lambda_{radio} = (3 \times 10^8 \text{ m/s}) / (1.42 \times 10^9 \text{ Hz}) \approx 0.21 m. Second, use the energy relationship to find the X-ray wavelength. Since energy is inversely proportional to wavelength (E∝1/λE \propto 1/\lambda), if the X-ray photon's energy is 10910^9 times greater, its wavelength must be 10910^9 times smaller. λxray=λradio/109=0.21 m/109=2.1×10−10\lambda_{xray} = \lambda_{radio} / 10^9 = 0.21 \text{ m} / 10^9 = 2.1 \times 10^{-10} m. This wavelength is in the X-ray part of the spectrum.

Question 10

An object emitting non-thermal synchrotron radiation is observed to have its peak emission shift from the radio band into the X-ray band. Which of the following physical changes would best explain this observation?

  1. The object's temperature increased dramatically, causing its blackbody peak to shift.
  2. The object moved closer to Earth, resulting in a significant blueshift of its spectrum.
  3. The charged particles producing the radiation were accelerated to much higher energies. (correct answer)
  4. The density of the gas surrounding the object increased, absorbing the radio waves.
Explanation: Synchrotron radiation is produced by relativistic charged particles spiraling in magnetic fields. The energy of the emitted photons is directly related to the energy of the particles and the strength of the magnetic field. A shift from low-energy radio waves to high-energy X-rays indicates that the emitted photons have become much more energetic. This can be caused by the charged particles being accelerated to much higher relativistic speeds or moving through a much stronger magnetic field. An increase in temperature would be relevant for thermal (blackbody) radiation, not synchrotron radiation. Doppler shift and absorption do not typically account for such a drastic shift across the spectrum from radio to X-ray.

Question 11

A space telescope uses a filter that transmits photons with energies between 2.2 eV and 2.8 eV. An astronomer uses this telescope to observe a star. Which colors of light from the star will be most prominently detected? (Note: The product of Planck's constant and the speed of light, hchc, is approximately 1240 eV·nm).

  1. A band in the red and orange part of the spectrum.
  2. A band in the blue and green part of the spectrum. (correct answer)
  3. Only ultraviolet light, as these energies are too high for visible light.
  4. The entire visible spectrum, from violet to red.
Explanation: First, convert the energy range to a wavelength range using the formula λ=hc/E\lambda = hc/E. The minimum wavelength is λmin=1240 eV⋅nm/2.8 eV≈443\lambda_{min} = 1240 \text{ eV}\cdot\text{nm} / 2.8 \text{ eV} \approx 443 nm. The maximum wavelength is λmax=1240 eV⋅nm/2.2 eV≈564\lambda_{max} = 1240 \text{ eV}\cdot\text{nm} / 2.2 \text{ eV} \approx 564 nm. This wavelength range of approximately 443 nm to 564 nm corresponds to the violet, blue, and green parts of the visible spectrum. Among the choices, 'blue and green' is the best description.

Question 12

An astronomer wants to create a multi-wavelength composite image of a nearby spiral galaxy. Which of the following correctly pairs a galactic component with the region of the electromagnetic spectrum best suited to reveal it?

  1. Cold neutral hydrogen gas in the spiral arms; Ultraviolet band.
  2. The population of old, cool stars in the central bulge; X-ray band.
  3. Dust lanes where new stars are forming; Infrared band. (correct answer)
  4. The supermassive black hole at the center; Visible light band.
Explanation: Different components of a galaxy shine brightest in different parts of the EM spectrum. Cold dust absorbs visible and UV light from young stars and re-radiates that energy in the infrared, making infrared observations ideal for tracing star-forming regions. (A) is incorrect because cold hydrogen gas is observed in the radio band (21-cm line). (B) is incorrect because old, cool stars emit mostly in visible and near-infrared light, not high-energy X-rays. (D) is incorrect because the black hole itself does not emit light, and the hot accretion disk around it is best observed in X-rays, not visible light.

Question 13

A spectral line from a distant galaxy is observed to have a significant redshift. How does the energy of the photons received at Earth compare to the energy of the photons as they were originally emitted by the galaxy, and what is the reason for this difference?

  1. The received photons have higher energy because they have gained kinetic energy during their journey through space.
  2. The received photons have lower energy because the expansion of space has stretched their wavelength. (correct answer)
  3. The received photons have the same energy because the energy of an individual photon is a conserved quantity.
  4. The received photons have lower energy because the galaxy's gravitational field trapped the more energetic photons.
Explanation: Cosmological redshift is the stretching of the wavelength of light due to the expansion of space. According to the relationship E=hc/λE = hc/\lambda, energy is inversely proportional to wavelength. As the wavelength increases (is redshifted), the energy of each photon decreases. The other options describe incorrect physical principles.

Question 14

The energy EE of a photon is related to its frequency ν\nu by E=hνE=h\nu, and its wavelength λ\lambda is related to its frequency by c=λνc=\lambda\nu. Which of the following expressions correctly solves for Planck's constant, hh, in terms of photon energy EE, wavelength λ\lambda, and the speed of light cc?

  1. h=Ec/λh = Ec/\lambda
  2. h=c/(Eλ)h = c/ (E\lambda)
  3. h=E/(λc)h = E/(\lambda c)
  4. h=Eλ/ch = E\lambda/c (correct answer)
Explanation: This question tests your ability to manipulate fundamental equations in electromagnetic radiation theory. When you encounter problems involving photon energy, frequency, and wavelength, you're working with two key relationships that describe how light behaves as both a wave and a particle. To solve for Planck's constant hh, you need to combine the given equations strategically. Start with E=hνE = h\nu and solve for hh: h=E/νh = E/\nu. Now you need to express frequency ν\nu in terms of the given variables. From c=λνc = \lambda\nu, you can solve for frequency: ν=c/λ\nu = c/\lambda. Substituting this into your expression for hh: h=E/(c/λ)=Eλ/ch = E/(c/\lambda) = E\lambda/c. Looking at the wrong answers: Option A (h=Ec/λh = Ec/\lambda) incorrectly multiplies rather than divides by the wavelength term, which would give units that don't match Planck's constant. Option B (h=c/(Eλ)h = c/(E\lambda)) has cc in the numerator instead of EE, fundamentally misrepresenting which quantities are proportional. Option C (h=E/(λc)h = E/(\lambda c)) places both λ\lambda and cc in the denominator, which comes from incorrectly substituting ν=λc\nu = \lambda c instead of ν=c/λ\nu = c/\lambda. The correct answer is D: h=Eλ/ch = E\lambda/c. When manipulating physics equations, always work step-by-step through substitutions and check that your final expression has the correct units. Planck's constant has units of energy × time, and you can verify that Eλ/cE\lambda/c yields these units.

Question 15

Why are space-based telescopes, such as the Chandra X-ray Observatory and the Fermi Gamma-ray Space Telescope, essential for making observations in their respective bands of the electromagnetic spectrum?

  1. The Earth's magnetic field bends the paths of high-energy photons, distorting images for ground-based observatories.
  2. Artificial light pollution on Earth is most intense at X-ray and gamma-ray wavelengths, overwhelming faint celestial sources.
  3. The Earth's atmosphere is opaque to most X-rays and gamma rays, absorbing them before they can reach the ground. (correct answer)
  4. High-energy photons cause significant atmospheric turbulence, which blurs images more severely than it does for visible light.
Explanation: The Earth's atmosphere effectively absorbs most electromagnetic radiation except for visible light, some infrared, and radio waves. High-energy radiation, such as X-rays and gamma rays, interacts with atoms and molecules in the upper atmosphere and is completely absorbed. Therefore, to observe the universe at these wavelengths, telescopes must be placed in orbit above the atmosphere. The other options are incorrect: Earth's magnetic field affects charged particles, not photons; light pollution is a problem in the visible spectrum; and atmospheric turbulence ('seeing') affects visible and near-infrared light, not high-energy radiation in the same way.

Question 16

An astronomical detector receives two signals from an emission nebula. The first signal consists of 120 photons per second of green light (λ≈500\lambda \approx 500 nm). The second signal consists of 100 photons per second of violet light (λ≈400\lambda \approx 400 nm). Which signal is depositing more total energy per second on the detector?

  1. The green light, because the detector is receiving a higher number of photons per second.
  2. The violet light, because each violet photon is significantly more energetic than each green photon. (correct answer)
  3. Both signals deposit the same amount of energy, as the higher photon count of the green light is balanced by the higher energy of the violet photons.
  4. It cannot be determined without knowing the efficiency of the detector at these wavelengths.
Explanation: The total energy per second (power) is the number of photons per second multiplied by the energy per photon. Photon energy is E=hc/λE=hc/\lambda. We need to compare Powergreen=120×(hc/500)Power_{green} = 120 \times (hc/500) with Powerviolet=100×(hc/400)Power_{violet} = 100 \times (hc/400). We can ignore the constant hchc and compare the ratios: For green light, the factor is 120/500=0.24120/500 = 0.24. For violet light, the factor is 100/400=0.25100/400 = 0.25. Since 0.25>0.240.25 > 0.24, the violet light signal is depositing slightly more energy per second, despite having fewer photons.

Question 17

The Cosmic Microwave Background (CMB) is thermal radiation that fills the universe. Its spectrum peaks at a wavelength of approximately 1.1 mm. Which statement accurately describes the CMB based on this information?

  1. The peak is in the infrared region, and its photons are more energetic than visible light photons.
  2. The peak is in the radio region, and its photons are less energetic than visible light photons.
  3. The peak is in the microwave region, and its photons are significantly less energetic than visible light photons. (correct answer)
  4. The peak is in the microwave region, and its photons are significantly more energetic than gamma-ray photons.
Explanation: A wavelength of 1.1 mm (1.1×10−31.1 \times 10^{-3} m) falls into the microwave portion of the electromagnetic spectrum. Visible light has wavelengths of approximately 400-700 nm (4−7×10−74-7 \times 10^{-7} m). Since the CMB's wavelength is about a thousand times longer than that of visible light, its photons are correspondingly a thousand times less energetic, as energy is inversely proportional to wavelength (E=hc/λE=hc/\lambda). Gamma rays have the shortest wavelengths and highest energies, so CMB photons are far less energetic.

Question 18

A spectral line from a distant galaxy is observed to have a significant redshift. How does the energy of the photons received at Earth compare to the energy of the photons as they were originally emitted by the galaxy, and what is the reason for this difference?

  1. The received photons have higher energy because they have gained kinetic energy during their journey through space.
  2. The received photons have lower energy because the expansion of space has stretched their wavelength. (correct answer)
  3. The received photons have the same energy because the energy of an individual photon is a conserved quantity.
  4. The received photons have lower energy because the galaxy's gravitational field trapped the more energetic photons.
Explanation: Cosmological redshift is the stretching of the wavelength of light due to the expansion of space. According to the relationship E=hc/λE = hc/\lambda, energy is inversely proportional to wavelength. As the wavelength increases (is redshifted), the energy of each photon decreases. The other options describe incorrect physical principles.

Question 19

An astronomer is designing an observatory to study the accretion disks around supermassive black holes, where matter is heated to temperatures exceeding 10 million Kelvin. Which region of the electromagnetic spectrum is most critical for this observatory to be sensitive to?

  1. Infrared, because accretion disks are often obscured by dust which glows at these wavelengths.
  2. Radio, because the strong magnetic fields in the disk produce intense synchrotron radiation.
  3. Ultraviolet, because this traces the hot, young stars that are frequently found near galactic centers.
  4. X-ray, because the extreme temperatures of the accreting gas cause it to emit high-energy thermal radiation. (correct answer)
Explanation: Objects heated to millions of Kelvin, such as the gas in an accretion disk around a black hole, emit thermal radiation that peaks at very short wavelengths. According to Wien's displacement law, such extreme temperatures correspond to peak emission in the X-ray portion of the electromagnetic spectrum. While some radio and infrared emission may be present, the defining characteristic of the hottest parts of the disk is X-ray emission.

Question 20

The spectrum of a hot, massive star peaks at a wavelength of 250 nm in the ultraviolet. If the star cools significantly over its lifetime so that its peak emission wavelength shifts to 500 nm in the visible spectrum, how does the energy of a typical photon emitted at the new peak wavelength compare to the energy of a typical photon from the original peak wavelength?

  1. The new peak photon energy is one-half of the original. (correct answer)
  2. The new peak photon energy is one-fourth of the original.
  3. The new peak photon energy is twice the original.
  4. The new peak photon energy is the same as the original.
Explanation: The energy of a photon is inversely proportional to its wavelength (E=hc/λE = hc/\lambda). The question asks about the energy of the photons at the peak wavelength, not the total energy output of the star. The original peak wavelength is λ1=250\lambda_1 = 250 nm. The new peak wavelength is λ2=500\lambda_2 = 500 nm. Since the wavelength has doubled (λ2=2λ1\lambda_2 = 2\lambda_1), the energy of a photon at this new peak must be halved (E2=E1/2E_2 = E_1 / 2). The distractor involving 'one-fourth' relates to the star's total luminosity (proportional to T4T^4), not individual photon energy.