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 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?

Short answer

(a) \( T_1 = T \sqrt{3} \); (b) Ratio is \( \frac{1}{\sqrt{3}} \).

Step-by-step solution

Understand the Stefan-Boltzmann Law

The Stefan-Boltzmann Law states that the power radiated per unit area from a blackbody is proportional to the fourth power of its temperature: \( P = \sigma A T^4 \).Where \( \sigma \) is the Stefan-Boltzmann constant, \( A \) is the surface area, and \( T \) is the temperature.

Express the Power for Each Star

Let \( T_1 \) and \( T_2 = T \) be temperatures of the hotter and cooler stars, respectively. The radius of the cooler star is 3 times the radius of the hotter star, so the surface area is \( A_2 = 9 A_1 \). Thus, the power emitted from each star can be given as follows:- For the hotter star: \( P_1 = \sigma A_1 T_1^4 \)- For the cooler star: \( P_2 = \sigma A_2 T_2^4 \).Since \( A_2 = 9 A_1 \), then \( P_2 = \sigma \times 9A_1 \times T^4 \).

Equate the Powers

Since both stars radiate the same total energy per second, we have \( P_1 = P_2 \). Substitute the expressions for \( P_1 \) and \( P_2 \):\[ \sigma A_1 T_1^4 = \sigma \times 9A_1 \times T^4 \].Cancel \( \sigma A_1 \) from both sides to simplify:\[ T_1^4 = 9T^4 \].

Solve for the Temperature of the Hotter Star

Taking the fourth root on both sides of the equation \( T_1^4 = 9T^4 \), we get:\[ T_1 = T \times 9^{1/4} \].Simplify this to find:\[ T_1 = T \times 3^{1/2} \]. This means \( T_1 = T \sqrt{3} \).So the temperature of the hotter star in terms of \( T \) is \( T \sqrt{3} \).

Understand Wien’s Displacement Law

Wien's Displacement Law states that the peak wavelength \( \lambda_{max} \) is inversely proportional to the temperature:\( \lambda_{max} = \frac{b}{T} \), where \( b \) is Wien's displacement constant.

Calculate the Ratio of Peak Wavelengths

Let \( \lambda_{1} \) and \( \lambda_{2} \) be the peak-intensity wavelengths for the hotter and cooler stars, respectively. From Wien’s Law, we have:\( \lambda_{1} = \frac{b}{T_1} \) and \( \lambda_{2} = \frac{b}{T} \).Thus, the ratio of the wavelengths is:\[ \frac{\lambda_{1}}{\lambda_{2}} = \frac{T}{T_1} = \frac{T}{T \sqrt{3}} = \frac{1}{\sqrt{3}} \].The ratio of the peak-intensity wavelength of the hot star to the peak-intensity wavelength of the cool star is \( \frac{1}{\sqrt{3}} \).

Key concepts

Stefan-Boltzmann Law

The Stefan-Boltzmann Law is a fundamental principle in the study of blackbody radiation. It describes how an ideal blackbody emits energy. According to this law, the power emitted by a blackbody per unit area is directly proportional to the fourth power of its temperature. This can be mathematically represented as:
\[P = \sigma A T^4\]
where:

• \( P \) is the power emitted,
• \( \sigma \) is the Stefan-Boltzmann constant, approximately \( 5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4 \),
• \( A \) is the surface area of the blackbody, and
• \( T \) is the temperature in Kelvin.

For two stars emitting the same power, the law dictates that a larger star must have a lower temperature if their radii are different. This is because the power output is also a function of the surface area. In our exercise, the cooler star, which is larger, must emit more from a larger area to match the same total power as the smaller hot star.

Wien's Displacement Law

Wien's Displacement Law provides a critical insight into understanding the behavior of blackbody radiation in terms of temperature and wavelength. This law states that the peak wavelength of the emitted radiation from a blackbody is inversely proportional to its temperature. The formula for Wien's Displacement Law is:
\[\lambda_{max} = \frac{b}{T}\]
where:

• \( \lambda_{max} \) is the peak wavelength,
• \( b \) is Wien's displacement constant, approximately \( 2.898 \times 10^{-3} \text{ m K} \), and
• \( T \) is the temperature in Kelvin.

In essence, as the temperature of a blackbody increases, the wavelength at which it emits most prominently shifts toward shorter wavelengths. Hence, hotter stars will have peak emissions in shorter wavelengths, meaning they appear bluer. Conversely, cooler stars will peak at longer wavelengths, often displaying a reddish color. This principle helps astronomers determine stellar temperatures based on their color.

Peak-intensity wavelength ratio

The concept of the peak-intensity wavelength ratio is derived from the application of Wien's Displacement Law, especially useful when comparing two blackbodies. From the exercise, we deal with two stars at different temperatures. Using Wien's Law, we can express the peak-intensity wavelengths for both stars given their temperatures.
For the hotter star:\[\lambda_1 = \frac{b}{T_1}\]
For the cooler star:\[\lambda_2 = \frac{b}{T}\]
This allows us to calculate the ratio of their peak wavelengths:
\[\frac{\lambda_1}{\lambda_2} = \frac{T}{T_1}\]
From our exercise, with the temperature of the hotter star being \( T \sqrt{3} \), the ratio simplifies to:\[\frac{1}{\sqrt{3}}\]
This tells us that the peak wavelength of the hotter star is shorter than that of the cooler star, precisely by the factor \( \sqrt{3} \). Such ratios are crucial in astrophysics, as they enable the understanding of the relative temperatures of celestial bodies through observational data.