Problem 48 - University Physics with Modern Physics

Question

Sirius B. The brightest star in the sky is Sirius, the Dog Star. It is actually a binary system of two stars, the smaller one (Sirius B) being a white dwarf. Spectral analysis of Sirius B indicates that its surface temperature is \(24,000 \mathrm{K}\) and that it radiates energy at a total rate of \(1.0 \times 10^{25} \mathrm{W}\) . Assume that it behaves like an ideal blackbody. (a) What is the total radiated intensity of Sirius \(\mathrm{B}\) (b) What is the peak-intensity wavelength? Is this wave-length visible to humans? (c) What is the radius of Sirius B? Express your answer in kilometers and as a fraction of our sun's radius. (d) Which star radiates more total energy per second, the hot Sirius or the (relatively) cool sun with a surface temperature of 5800 \(\mathrm{K}\) ? To find out, calculate the ratio of the total power radiated by our sun to the power radiated by Sirius B.

Step-by-Step Solution

Answer

(a) Total intensity: \(2.8 \times 10^{13} \text{ W/m}^2\). (b) Peak wavelength: 121 nm (ultraviolet). (c) Radius: 5800 km; 0.83% of Sun's radius. (d) Sun radiates 38 times more energy.

Step 1: Total Radiated Intensity

To find the total radiated intensity, we use the Stefan-Boltzmann Law. Intensity \(I\) is given by \( I = \frac{P}{A} \), where \(P\) is the total power radiated and \(A\) is the surface area. Sirius B is a sphere, so its surface area \(A\) is \(4\pi R^2\). However, we are given \(P = 1.0 \times 10^{25} \text{ W}\), so the total radiated intensity \(I\) of Sirius B is: \[ I = \sigma T^4 = \frac{P}{A} \] where \(\sigma\) is the Stefan-Boltzmann constant \(5.67 \times 10^{-8} \text{ W/m}^2 \text{ K}^4\). Thus, \[ I = \sigma (24000)^4 \]. Calculating this gives \[ I \approx 2.8 \times 10^{13} \text{ W/m}^2 \].

Step 2: Peak-Intensity Wavelength

Use Wien's Displacement Law: \( \lambda_{\text{max}} = \frac{b}{T} \) where \(b\) is Wien's displacement constant \(2.9 \times 10^{-3} \text{ m K}\). So, \( \lambda_{\text{max}} = \frac{2.9 \times 10^{-3}}{24000} \approx 1.21 \times 10^{-7} \text{ m} \) or 121 nm. This wavelength is in the ultraviolet range, which is not visible to humans.

Step 3: Calculate Radius of Sirius B

We rearrange the Stefan-Boltzmann Law to find \(R\), where \(P = \sigma T^4 (4\pi R^2)\). Solving for \(R\), we have: \[ R = \sqrt{\frac{P}{4\pi \sigma T^4}} \]. Substituting the given values, \(R = \sqrt{\frac{1.0 \times 10^{25}}{4\pi (5.67 \times 10^{-8}) (24000)^4}} \approx 5.8 \times 10^6 \text{ m} \). Convert to kilometers: \(5.8 \times 10^3 \text{ km}\). Assuming our Sun's radius is \(7.0 \times 10^5 \text{ km}\), Sirius B's radius is about \(0.0083\) or \(0.83\%\) of the Sun's radius.

Step 4: Compare Total Energy Radiated

Let \(P_\text{sun}\) and \(P_\text{Sirius B}\) be the powers radiated by the Sun and Sirius B, respectively. Using the Stefan-Boltzmann Law: \(P_\text{sun} = \sigma T_\text{sun}^4 (4\pi R_\text{sun}^2)\). Given \(T_\text{sun} = 5800\, \text{K}\), we calculate the ratio: \[ \frac{P_\text{sun}}{P_\text{Sirius B}} = \frac{\sigma (5800)^4 (4\pi (7 \times 10^8)^2)}{1.0 \times 10^{25}} \]. \( P_\text{sun} \approx 3.8 \times 10^{26} \text{ W} \), giving a ratio \( \approx 38 \). Thus, the Sun radiates about 38 times more energy than Sirius B.

Key Concepts

Wien's Displacement Law

Wien's Displacement Law is an important principle in the study of blackbody radiation. It tells us how the temperature of an ideal blackbody affects the wavelength at which it emits the most potent radiation. In simpler terms, Wien's Displacement Law provides a way to find the peak wavelength of emission for any object that behaves like a blackbody, depending on its temperature.**How it Works**
- The law is represented by the equation \( \lambda_{\text{max}} = \frac{b}{T} \). Here, \( \lambda_{\text{max}} \) denotes the wavelength at which the radiation is strongest, \( T \) is the temperature of the object, and \( b \) is Wien's constant, approximately \( 2.9 \times 10^{-3} \text{ m K} \).
- This tells us that as the temperature of a blackbody increases, the peak emission wavelength shifts to shorter wavelengths. For example, a hotter object will emit its peak radiation in the ultraviolet range, while a cooler object might emit in the infrared.**Example Application: Sirius B**
- In the case of Sirius B, with a surface temperature of 24,000 K, Wien's Law helps us calculate its peak intensity wavelength as about 121 nm, which is in the ultraviolet range, making it invisible to the human eye.

blackbody radiation

Blackbody radiation refers to the theoretical perfect emission of energy from an object that absorbs all incident light, making it a perfect emitter. Such an ideal blackbody does not reflect or transmit any light.**Essential Characteristics**
- A blackbody in thermal equilibrium emits radiation called "blackbody radiation," which only depends on the body's temperature, not its shape or material composition.- This radiation covers all wavelengths and follows Planck's Law, which describes the spectral energy distribution of radiation emitted by a blackbody.**Stefan-Boltzmann Law**
- One of the key laws related to blackbody radiation is the Stefan-Boltzmann Law, which quantifies the total energy emitted by a blackbody.- The formula is \( P = \sigma T^4 A \), where \( P \) is the power output, \( \sigma \) is the Stefan-Boltzmann constant, \( T \) is the absolute temperature in Kelvin, and \( A \) is the surface area of the blackbody.
- This law shows that the energy radiated by a blackbody increases with the fourth power of its temperature, meaning a small increase in temperature leads to a significant increase in energy emitted.**Application: Sirius B**
- Sirius B, assumed as an ideal blackbody, radiates energy based on its temperature, allowing us to calculate parameters like its total radiated intensity, peak-intensity wavelength, and its physical dimensions through this concept.

white dwarf stars

White dwarf stars are fascinating remnants of stellar evolution. They are the last observable stage of medium to small stars, including stars like our Sun when they exhaust their nuclear fuel. **Characteristics**
- White dwarfs are incredibly dense, meaning they pack a lot of mass into a tiny volume, resulting in very high surface temperatures.
- They are typically about the size of Earth, but possess a mass comparable to the Sun, due to this immense density. - A white dwarf does not generate energy by nuclear reactions, unlike normal stars. Instead, it glows because of leftover thermal energy. **Connection to Blackbody Radiation**
- A white dwarf like Sirius B behaves similarly to an ideal blackbody, emitting radiation that can be analyzed using the laws of blackbody radiation. - The high temperature of a white dwarf results in radiation emitted in the ultraviolet and visible spectra, shedding light on their physical properties when studied through spectral analysis. **Role in Astronomy**
- Studying white dwarfs helps astronomers understand cooling rates, life cycles of stars, and the history of stellar formation in the universe. Their unique physical properties make them important for testing theories of physics under extreme conditions.