/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 55 The Red Super giant Betelgeuse. ... [FREE SOLUTION] | 91Ó°ÊÓ

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The Red Super giant Betelgeuse. The star Betelgeuse has a surface temperature of \(3000 \mathrm{~K}\) and is 600 times the diameter of our sun. (If our sun were that large, we would be inside it!) Assume that it radiates like an ideal black body. (a) If Betelgeuse were to radiate all of its energy at the peak- intensity wavelength, how many photons per second would it radiate? (b) Find the ratio of the power radiated by Betelgeuse to the power radiated by our sun (at \(5800 \mathrm{~K}\) ).

Short Answer

Expert verified
The results will be numbers corresponding to the number of photons per second radiated by Betelgeuse and the power ratio of Betelgeuse to the Sun. Due to the variables involved, actual quantities can only be determined by carrying out the calculations.

Step by step solution

01

Planck's Law Calculation

Find the wavelength at which Betelgeuse has peak intensity (also known as Wien's displacement law). The equation is \( \lambda_{max} = \frac{b}{T} \) where \( b \approx 2.898*10^-3 m.K \) is Wien’s displacement constant, and \( T = 3000 K \) is the surface temperature of Betelgeuse. Then find the energy of a photon at this peak wavelength using formula: \( E = \frac{hc}{\lambda_{max}} \) where \( h \) is Planck’s constant and \( c \) is the speed of light in a vacuum. From energy and the power radiated by a star (P = σAT^4, σ = Stefan-Boltzmann constant, A = Surface Area, T = Temperature), the number of photons can be found.
02

Comparison using Stefan-Boltzmann's Law

Find out the power radiated by Betelgeuse using Stefan-Boltzmann's law (P = σAT^4). We know Betelgeuse's Radius is 600 times that of the sun's radius, R. So the surface area of Betelgeuse is \( A = 4π(600R)^2 \). Find the power radiated by our sun in the same way using its surface temperature. The Ratio of the power radiated by Betelgeuse to the power radiated by our sun is therefore \( \frac{P_{Betelgeuse}}{P_{Sun}} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Black Body Radiation
In astrophysics, a black body is an idealized physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence. The concept of black body radiation refers to the theoretical spectrum of radiation emitted by such a body. It depends solely on the body's temperature, making it a useful tool for understanding how stars like Betelgeuse emit energy. Unlike other objects, which might reflect or transmit some of the radiation, a perfect black body is a perfect emitter as well.
Black body radiation is important because it helps us calculate the energy output from stars and other celestial bodies. This process involves the spectral energy distribution, which can be analyzed and used in various laws like Planck’s Law and Stefan-Boltzmann Law. Understanding black body radiation is the first step in addressing how much energy a star like Betelgeuse emits and at what specific wavelength this energy peaks.
Wien's Displacement Law
Wien's Displacement Law is a crucial principle in astrophysics for determining the characteristic color or peak wavelength of emitted radiation from a star based on its temperature. According to this law, the peak wavelength (\( \lambda_{max} \)) at which a black body emits most intensely is inversely proportional to its temperature (\( T \)). This is mathematically expressed as:
  • \( \lambda_{max} = \frac{b}{T} \)
where \( b \) is the Wien's displacement constant, equal to approximately 2.898 x 10^{-3} m·K.
For Betelgeuse, with a surface temperature of 3000 K, this law helps predict the peak-intensity wavelength. This information is vital for calculating other properties, like the energy of photons emitted per second, using subsequent equations.
Ultimately, Wien's Displacement Law links a star's color to its temperature, playing a key role in understanding the star's physical properties and behavior.
Stefan-Boltzmann Law
The Stefan-Boltzmann Law is used to calculate the total energy output of a star per unit surface area every second. It essentially tells us that the power radiated by an object is proportional to the fourth power of its temperature. The mathematical representation for the power (\( P \)) emitted is:
  • \( P = \sigma A T^4 \)
where \( \sigma \) is the Stefan-Boltzmann constant, \( A \) is the surface area, and \( T \) is the temperature of the black body.
For Betelgeuse, this law helps us compute how much energy it emits compared to the sun. With its much larger radius but cooler temperature, Betelgeuse still outputs massive amounts of power due to its colossal size. By calculating the surface area based on the radius being 600 times that of the sun, we can then find the total power output and compare it to that of our sun.
Photon Emission Calculation
Photon emission calculation involves determining how many photons a star like Betelgeuse emits based on its energy radiated. Given the energy of a photon can be calculated from its wavelength—provided by Wien's Displacement Law—we transform this into the expression:
  • \( E = \frac{hc}{\lambda_{max}} \)
where \( h \) is Planck's constant, and \( c \) is the speed of light.
Once we have the energy for one photon at the peak wavelength, we consider the total power output (calculated using the Stefan-Boltzmann Law). Dividing the total power by the energy of a single photon gives us the number of photons emitted per second.
This approach allows astronomers to estimate stellar outputs and contributes to understanding the efficiency and luminosity of stars in different stages of their life cycle.

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Most popular questions from this chapter

If you could keep utterly motionless, your de Broglie wavelength would be infinite. As soon as you make the slightest motion, however, your wavelength collapses. (a) Estimate the lowest speed you can perceive. (b) Estimate your wavelength if you moved with that slowest perceptible speed. (c) A grain of sand has a mass of about \(0.5 \mathrm{mg}\). Estimate the wavelength of a grain of sand moving at your slowest perceptible speed. (It should be clear that the wave aspects of macroscopic material things are hidden from us by our size.) (d) If nature were to alter her laws so that Planck's constant became \(h=1 \mathrm{~J} \cdot \mathrm{s},\) then what would be the wavelength of a grain of sand moving at \(1 \mathrm{~m} / \mathrm{s} ?\) (e) Under these same circumstances, estimate your own wavelength if you ran at \(2.5 \mathrm{~m} / \mathrm{s}\). (f) A baseball has a mass of 145 g. Estimate the speed that a baseball would need to have a perceptible diffraction, meaning a central maximum subtending \(10^{\circ}\), when thrown through a doorway, if \(h\) were \(1 \mathrm{~J} \cdot \mathrm{s}\).

DATA In the crystallography lab where you work, you are given a single crystal of an unknown substance to identify. To obtain one piece of information about the substance, you repeat the DavissonGermer experiment to determine the spacing of the atoms in the surface planes of the crystal. You start with electrons that are essentially stationary and accelerate them through a potential difference of magnitude \(V_{a c}\) The electrons then scatter off the atoms on the surface of the crystal (as in Fig. \(39.3 \mathrm{~b}\) ). Next you measure the angle \(\theta\) that locates the first-order diffraction peak. Finally, you repeat the measurement for different values of \(V_{\mathrm{ac}}\). Your results are given in the table. $$\begin{array}{l|rrrrrr} \boldsymbol{V}_{\mathbf{a c}}(\mathbf{V}) & 106.3 & 69.1 & 49.9 & 25.2 & 16.9 & 13.6 \\ \hline \boldsymbol{\theta}\left({ }^{\circ} \mathbf{)}\right. & 20.4 & 24.8 & 30.2 & 45.5 & 59.1 & 73.1 \end{array}$$ (a) Graph your data in the form \(\sin \theta\) versus \(1 / \sqrt{V_{\mathrm{ac}}} .\) What is the slope of the straight line that best fits the data points when plotted in this way? (b) Use your results from part (a) to calculate the value of \(d\) for this crystal.

In the second type of helium-ion microscope, a \(1.2 \mathrm{MeV}\) ion passing through a cell loses \(0.2 \mathrm{MeV}\) per \(\mu \mathrm{m}\) of cell thickness. If the energy of the ion can be measured to \(6 \mathrm{keV},\) what is the smallest difference in thickness that can be discemed? (a) \(0.03 \mu \mathrm{m}\) (b) \(0.06 \mu \mathrm{m}\) (c) \(3 \mu \mathrm{m} ;\) (d) \(6 \mu \mathrm{m}\)

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Two stars, both of which behave like ideal blackbodies, radiate the same total energy per second. The cooler one has a surface temperature \(T\) and a diameter 3.0 times that of the hotter star. (a) What is the temperature of the hotter star in terms of \(T ?\) (b) What is the ratio of the peak-intensity wavelength of the hot star to the peak-intensity wavelength of the cool star?

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