Problem 46 - University Physics with Modern Physics

Question

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 3.0 times the diameter 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?

Step-by-Step Solution

Answer

(a) The temperature of the hotter star is \(T \sqrt{3}\). (b) The wavelength ratio is \(1/\sqrt{3}\).

Step 1: Understand the problem

We have two stars behaving as ideal blackbodies that radiate the same total energy per second. The cooler star has a surface temperature \(T\) and its diameter is 3.0 times that of the hotter star. We need to find the temperature of the hotter star and the ratio of their peak-intensity wavelengths.

Step 2: Apply Stefan-Boltzmann Law

The Stefan-Boltzmann Law states the total energy radiated per unit surface area of a blackbody is \(E = \sigma T^4\). The energy from the whole star is proportional to its surface area \(A = \pi D^2\) and can be expressed as \(E = \sigma T^4 \times \pi D^2\).

Step 3: Calculate the surface area ratio

Let \(D_1\) be the diameter of the hotter star and \(D_2 = 3D_1\) be the diameter of the cooler star. The surface area ratio is \(A_2/A_1 = (\pi (3D_1)^2) / (\pi D_1^2) = 9\).

Step 4: Set up the equation with equal energy outputs

Since both stars radiate the same energy, we equate their total energy output: \(\sigma T^4 \pi (3D_1)^2 = \sigma T_h^4 \pi D_1^2\). Simplifying gives \(9T^4 = T_h^4\).

Step 5: Solve for the temperature of the hotter star

We solve the equation \(9T^4 = T_h^4\) for \(T_h\): \(T_h^4 = 9T^4\), therefore \(T_h = T\sqrt[4]{9} = T\times \sqrt{3}\).

Step 6: Determine the ratio of peak-intensity wavelengths

By Wien's Displacement Law, \(\lambda_{peak} T = constant\). For two sources at temperatures \(T\) and \(T_h\), \(\frac{\lambda_{hot}}{\lambda_{cool}} = \frac{T}{T_h} = \frac{T}{T\sqrt{3}} = \frac{1}{\sqrt{3}}\).

Key Concepts

Stefan-Boltzmann Law

The Stefan-Boltzmann Law is vital in understanding how stars emit energy. It expresses that the power output of a blackbody, like a star, is proportional to the fourth power of its temperature. This is given by the formula:\[E = \sigma T^4 \]where \( E \) is the energy radiated per unit area, \( T \) is the surface temperature of the star, and \sigma is the Stefan-Boltzmann constant. This law highlights that even a small change in temperature results in a significant change in the energy output.Let's break it down a bit further:- The energy radiated is not only dependent on temperature but also the surface area of the star. Larger stars can emit more energy due to their greater surface area.- For our problem, even though two stars emit the same total energy, the cooler star must do so by compensating for its lower temperature with a larger diameter.Understanding this principle allows us to predict that the hotter star, being smaller, must have a higher temperature to match the energy output of the larger, cooler star.

Wien's Displacement Law

Wien’s Displacement Law reveals the relationship between the temperature of a blackbody and the wavelength at which it emits most intensely. It is expressed as:\[\lambda_{peak} \cdot T = b\]where \(\lambda_{peak}\) is the peak wavelength, \(T\) is the temperature, and \(b\) is Wien's constant.Here's how it helps:- As the temperature of a star increases, the peak wavelength shifts to shorter wavelengths. This means that hotter stars tend to emit light more in the blue/ultraviolet range, while cooler stars emit more in the red/infrared range.- Given two stars, we can determine the ratio of their peak wavelengths if their temperatures are known. In the exercise, this ratio is derived from the temperatures of the cooler and hotter stars.This principle helps astronomers determine the color and type of star they are observing by examining the light spectrum. For our stars, translating temperature differences into wavelength ratios explains why the hotter star is bluer than its cooler, redder counterpart.

Energy Output of Stars

Stars radiate energy across vast distances, with their output measured in terms of luminosity. This luminosity is the total energy emitted per second, usually in watts. Several factors play into this: - **Surface Temperature:** As explained by both the Stefan-Boltzmann Law and Wien's Law, temperature is a critical factor. - **Surface Area:** Larger stars can have higher luminosities due to their greater surface area, despite potentially lower temperature. In the context of our problem: - Both stars emit the same total energy per second. The larger, cooler star achieves this because its larger surface area compensates for its lower temperature. - The hotter star, though smaller, achieves the same energy output due to its higher temperature. This understanding of energy output helps astronomers not only measure the inherent brightness of stars but also compare them to each other and predict other characteristics, such as potential size and life cycle stages.