Question

Light from an ideal spherical blackbody 15.0 cm in diameter is analyzed by using a diffraction grating that has 3850 lines/cm. When you shine this light through the grating, you observe that the peak-intensity wavelength forms a first-order bright fringe at \(\pm\)14.4\(^\circ\) from the central bright fringe. (a) What is the temperature of the blackbody? (b) How long will it take this sphere to radiate 12.0 MJ of energy at constant temperature?

Short answer

(a) The temperature is approximately 4490 K. (b) It takes about 4.74 hours to radiate 12.0 MJ.

Step-by-step solution

Identify the Diffraction Grating Equation

The diffraction grating equation is given by \[ d \sin \theta = m \lambda \]where \( d \) is the distance between grating lines (grating spacing), \( \theta \) is the diffraction angle, \( m \) is the order of the fringe (here, \( m = 1 \) for first-order), and \( \lambda \) is the wavelength. We need to find \( d \) from the given lines per centimeter: \[ d = \frac{1}{3850 \text{ lines/cm}} = \frac{1}{385000 \text{ lines/m}} \]

Calculate the Wavelength

Substitute \( d \) and \( \theta = 14.4^\circ \) into the diffraction grating equation to find \( \lambda \):\[ \sin 14.4^\circ \approx 0.248 \]\[ \frac{1}{385000} \cdot 0.248 = 1 \cdot \lambda \]Solve for \( \lambda \):\[ \lambda \approx \frac{0.248}{385000} \approx 6.45 \times 10^{-7} \text{ m} \](645 nm).

Use Wien's Law to Find the Temperature

Wien's displacement law relates the peak wavelength \( \lambda_{max} \) to the temperature \( T \) of a blackbody:\[ \lambda_{max} T = b \]where \( b = 2.898 \times 10^{-3} \text{ m}\cdot\text{K} \).Substituting the wavelength:\[ 6.45 \times 10^{-7} \cdot T = 2.898 \times 10^{-3} \]Solve for \( T \):\[ T = \frac{2.898 \times 10^{-3}}{6.45 \times 10^{-7}} \approx 4.49 \times 10^3 \text{ K} \]

Calculate Power Using Stefan-Boltzmann Law

The power radiated by a spherical blackbody is given by:\[ P = \sigma A T^4 \]where \( \sigma = 5.67 \times 10^{-8} \text{ W/m}^2\cdot\text{K}^4 \), \( A \) is the surface area of the sphere \( A = 4\pi r^2 \), and \( r = 0.075 \text{ m} \) (half the diameter).Calculate the surface area:\[ A = 4\pi (0.075)^2 \approx 0.0707 \text{ m}^2\]Now calculate the power:\[ P = 5.67 \times 10^{-8} \times 0.0707 \times (4.49 \times 10^3)^4 \approx 703 \text{ W} \]

Find Time to Radiate 12.0 MJ

To find the time \( t \) it takes to radiate \( 12.0 \text{ MJ} \) with power \( P \), we use:\[ t = \frac{E}{P} \]where \( E = 12.0 \times 10^6 \text{ J} \), and \( P = 703 \text{ W} \).Substitute and solve for \( t \):\[ t = \frac{12.0 \times 10^6}{703} \approx 17070 \text{ s} \approx 4.74 \text{ hours} \]

Key concepts

Diffraction Grating

Diffraction grating is an optical component with a pattern of evenly spaced lines or grooves that diffract and disperse light into its components. These grating lines can be extremely small, determining how light waves scatter when they pass through or reflect off the grating. This phenomenon allows us to see the spectrum of different wavelengths that make up the light.

When light hits a diffraction grating, it bends at specific angles depending on the wavelength. This bending causes the light to produce what is known as diffraction lines or diffraction maxima. The equation governing this is:

• \[ d \sin \theta = m \lambda \]

Where:

• \( d \) represents the distance between the grating lines (grating spacing).
• \( \theta \) is the angle at which the maximum occurs.
• \( m \) is the order of the maximum (e.g., first, second order), typically an integer.
• \( \lambda \) is the wavelength of the light.

For example, in the provided exercise, a light of known wavelength forms a first-order maximum at a specific angle, allowing one to calculate the light’s wavelength using the number of lines per meter on the grating. This principle is crucial for separating different wavelengths in practical tools like spectrometers.

Wien's Displacement Law

Wien’s displacement law offers a relationship between the temperature of a blackbody and the peak wavelength of its emitted radiation. Named after Wilhelm Wien, this law allows scientists to determine the temperature of an object from the color of its emitted light. It is particularly useful in astronomy, helping in estimating the temperature of stars.

The formula for Wien's displacement law is:

• \[ \lambda_{max} T = b \]

Where:

• \( \lambda_{max} \) is the peak wavelength in meters.
• \( T \) is the absolute temperature in Kelvin.
• \( b \) is Wien’s displacement constant \( 2.898 \times 10^{-3} \text{ m} \cdot \text{K} \).

For instance, by knowing the peak wavelength of a blackbody, we can rearrange the formula to solve for temperature \( T \). In the exercise at hand, using the observed wavelength of light, we calculate the temperature of the blackbody, which comes out to be approximately 4490 K. This converts the study of spectra into a powerful tool for understanding thermal properties of various bodies.

Stefan-Boltzmann Law

The Stefan-Boltzmann law describes the power radiated per unit area of a blackbody in terms of its temperature. This law posits that the energy radiated by a perfect blackbody per unit area is directly proportional to the fourth power of its absolute temperature. This is a fundamental law in the field of thermodynamics.

Mathematically, it is represented as:

• \[ P = \sigma A T^4 \]

Where:

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

For a spherical body, the surface area \( A \) can be calculated using \( 4\pi r^2 \). In our example, once the surface area and temperature are known, the Stefan-Boltzmann law is used to find out how much energy the blackbody radiates per second. This knowledge helps us to calculate how long it would take for the blackbody to radiate a particular amount of energy, as in our exercise, finding it takes approximately 4.74 hours to radiate 12 MJ.