EBK LOOSE-LEAF VERSION OF UNIVERSE
11th Edition
ISBN: 9781319227975
Author: KAUFMANN
Publisher: VST
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Chapter 20, Problem 13CC
To determine
The reason behind SN 1987A about one-tenth as bright as initially expected.
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For the PP chain 0.7% of the mass participating in nuclear fusion is liberated as energy which
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?max =
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How does that compare with 91.2 nm, the wavelength of photons with just enough energy to ionize hydrogen?
-The wavelength calculated above is shorter than 91.2 nm. Photons at this calculated wavelength will have more than enough energy to ionize hydrogen.
-The wavelength calculated above is longer than 91.2 nm. Photons at this calculated wavelength will have more than enough energy to ionize hydrogen.
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-The wavelength calculated above is longer than 91.2 nm. Photons at this calculated wavelength will not have enough energy to ionize hydrogen.
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Chapter 20 Solutions
EBK LOOSE-LEAF VERSION OF UNIVERSE
Ch. 20 - Prob. 1CCCh. 20 - Prob. 2CCCh. 20 - Prob. 3CCCh. 20 - Prob. 4CCCh. 20 - Prob. 5CCCh. 20 - Prob. 6CCCh. 20 - Prob. 7CCCh. 20 - Prob. 8CCCh. 20 - Prob. 9CCCh. 20 - Prob. 10CC
Ch. 20 - Prob. 11CCCh. 20 - Prob. 12CCCh. 20 - Prob. 13CCCh. 20 - Prob. 14CCCh. 20 - Prob. 15CCCh. 20 - Prob. 16CCCh. 20 - Prob. 17CCCh. 20 - Prob. 18CCCh. 20 - Prob. 1QCh. 20 - Prob. 2QCh. 20 - Prob. 3QCh. 20 - Prob. 4QCh. 20 - Prob. 5QCh. 20 - Prob. 6QCh. 20 - Prob. 7QCh. 20 - Prob. 8QCh. 20 - Prob. 9QCh. 20 - Prob. 10QCh. 20 - Prob. 11QCh. 20 - Prob. 12QCh. 20 - Prob. 13QCh. 20 - Prob. 14QCh. 20 - Prob. 15QCh. 20 - Prob. 16QCh. 20 - Prob. 17QCh. 20 - Prob. 18QCh. 20 - Prob. 19QCh. 20 - Prob. 20QCh. 20 - Prob. 21QCh. 20 - Prob. 22QCh. 20 - Prob. 23QCh. 20 - Prob. 24QCh. 20 - Prob. 25QCh. 20 - Prob. 26QCh. 20 - Prob. 27QCh. 20 - Prob. 28QCh. 20 - Prob. 29QCh. 20 - Prob. 30QCh. 20 - Prob. 31QCh. 20 - Prob. 32QCh. 20 - Prob. 33QCh. 20 - Prob. 34QCh. 20 - Prob. 35QCh. 20 - Prob. 36QCh. 20 - Prob. 37QCh. 20 - Prob. 38QCh. 20 - Prob. 39QCh. 20 - Prob. 40QCh. 20 - Prob. 41QCh. 20 - Prob. 42QCh. 20 - Prob. 43QCh. 20 - Prob. 44QCh. 20 - Prob. 45QCh. 20 - Prob. 46QCh. 20 - Prob. 47QCh. 20 - Prob. 48QCh. 20 - Prob. 49QCh. 20 - Prob. 50QCh. 20 - Prob. 51QCh. 20 - Prob. 52QCh. 20 - Prob. 53QCh. 20 - Prob. 54QCh. 20 - Prob. 55QCh. 20 - Prob. 56QCh. 20 - Prob. 57QCh. 20 - Prob. 58QCh. 20 - Prob. 59QCh. 20 - Prob. 60QCh. 20 - Prob. 61QCh. 20 - Prob. 62QCh. 20 - Prob. 63QCh. 20 - Prob. 64QCh. 20 - Prob. 65QCh. 20 - Prob. 66QCh. 20 - Prob. 67QCh. 20 - Prob. 68QCh. 20 - Prob. 69QCh. 20 - Prob. 70QCh. 20 - Prob. 71QCh. 20 - Prob. 72QCh. 20 - Prob. 73QCh. 20 - Prob. 74QCh. 20 - Prob. 75Q
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- Before the star that became SN 1987A exploded, it evolved from a red supergiant to a blue supergiant while remaining at the same luminosity. As a red supergiant, its surface temperature would have been approximately 4000 K, while as a blue supergiant, its surface temperature was 16,000 K. How much did the radius change as it evolved from a red to a blue supergiant?arrow_forwardH II regions can exist only if there is a nearby star hot enough to ionize hydrogen. Hydrogen is ionized only by radiation with wavelengths shorter than 91.2 nm. What is the temperature of a star that emits its maximum energy at 91.2 nm? (Use Wien’s law from Radiation and Spectra.) Based on this result, what are the spectral types of those stars likely to provide enough energy to produce H II regions?arrow_forwardWhat is the acceleration of gravity at the surface of the star that became SN 1987A? How does this g compare to that at the surface of Earth? The mass was 20 times that of the Sun and the radius was 41 times that of the Sun.arrow_forward
- What was the average density of the star that became SN 1987A? How does it compare to the average density of Earth? The mass was 20 times that of the Sun and the radius was 41 times that of the Sun.arrow_forwardWhy do nebulae near hot stars look red? Why do dust clouds near stars usually look blue?arrow_forwardIf the Sun were a member of the cluster NGC 2264, would it be on the main sequence yet? Why or why not?arrow_forward
- Betelgeuse is a nearby supergiant that will eventually explode into a supernova. Let's see how awesome it would look. At peak brightness, the supernova will have a luminosity of about 10 billion times the Sun. It is 600 light-years away. All stellar brightnesses are compared with Vega, which has an intrinsic luminosity of about 60 times the Sun, a distance of 25 light-years, an absolute magnitude of 0.6 and an apparent magnitude of 0 (by definition). a) At peak brightness, how many times brighter will Betelgeuse be than Vega? b) Approximately what apparent magnitude does this correspond to? c) The Sun is about -26.5 apparent magnitude. What fraction of the Sun's brightness will Betelgeuse be?arrow_forward1 Solar constant, Sun, and the 10 pc distance! The luminosity of Sun is + 4- 1026 W - 4- 1033ergs-1, The Sun is located at a distance of m from the Earth. The Earth receives a radiant flux (above its atmosphere) of F = 1365W m- 2, also known as the solar constant. What would have been the Solar contact if the Sun was at a distance of 10 pc ? 1AU 1 1.5-+ 1011arrow_forwardThe apparent magnitude of a star is observed to vary between m = +0.4 and m = +0.1 because the star pulsates and hence continuously changes its radius and temperature. When at its peak brightness, the star’s radius has increased by a factor of two compared to its value at the mini- mum brightness. Determine the value of T+/T−, where T+ is the temperature when the star is at its peak brightness and T− is the temperature when the star is at it minimum brightness. Note: we expect T+/T− < 1 because the star’s temperature decreases as its radius increases.arrow_forward
- Asap plzzzarrow_forwardAn O8 V star has an apparent visual magnitude of +5. Use the method of spectroscopic parallax to estimate the distance to the star (in pc). (Hints: Refer to one of the H–R diagrams in the chapter, and use the magnitude–distance formula, d = 10(mV − MV + 5)/5 where d is the distance in parsecs, mV and MV are the apparent and absolute visual magnitude respectively.)arrow_forwardIf the surface Temperature of a star was about 11000.0 K instead of 7000.0 K what is the ratio of power per square meter of the 11000.0 K star compared to power per square meter of the 7000.0 K star? How many times greater is the magnitude of power per square meter of the 11000.0 K star compared to the 7000.0 K stararrow_forward
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