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Chapter 39: Problem 62

Light from an ideal spherical blackbody 15.0 \(\mathrm{cm}\) in diameter is analyzed using a diffraction grating having 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 \(\pm 11.6^{\circ}\) 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 \(\mathrm{MJ}\) of energy?

Short Answer

Expert verified
(a) The temperature is approximately 4929 K. (b) The time taken to radiate 12.0 MJ of energy is about 1.2 hours.

Step by step solution

01

Calculate the Wavelength using Diffraction Formula

Using the grating equation for first-order diffraction, \( d \sin \theta = m \lambda \), where \(d\) is the grating spacing, \(\theta\) is the angle, \(m\) is the order (which is 1 in this case), and \(\lambda\) is the wavelength. The grating spacing is \(d = \frac{1}{3850 \, \text{lines/cm}}\), converted to meters, gives \(d = \frac{1}{385000 \, \text{lines/m}}\). With \(\theta = 11.6^{\circ}\), calculate the wavelength \( \lambda \).
02

Use Wien's Displacement Law to Find Temperature

Wien鈥檚 displacement law is \( \lambda_{max} T = b \), where \( b = 2.8977719 \times 10^{-3} \, \text{m} \cdot \text{K} \). Rearrange to find the temperature \( T = \frac{b}{\lambda_{max}} \). Substitute the wavelength \( \lambda \) found from Step 1 to find the temperature of the blackbody.
03

Calculate Surface Area of the Blackbody

The surface area \(A\) of a sphere is given by \(A = 4 \pi r^2\). Given the diameter of the sphere is 15.0 cm, calculate the radius \(r = 7.5 \, \text{cm} = 0.075 \, \text{m}\), and then calculate the surface area.
04

Use Stefan-Boltzmann Law to Find Power Output

The Stefan-Boltzmann law states that the power \( P \) radiated by a blackbody is \( P = \sigma A T^4 \), where \( \sigma = 5.67 \times 10^{-8} \, \text{W/m}^2\cdot\text{K}^4 \). Calculate the power using the area from Step 3 and temperature from Step 2.
05

Calculate Time to Radiate Required Energy

The energy \( E \) radiated is related to power by \( E = P t \), where \( t \) is the time. Given \( E = 12.0 \times 10^6 \, \text{J} \), solve for time: \( t = \frac{E}{P} \). Use the power calculated in Step 4 to find the time.

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Key Concepts

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

Diffraction Grating
A diffraction grating is a tool used to disperse light into its component wavelengths. It functions by using a surface etched or ruled with many closely spaced lines, such as 3850 lines per centimeter as mentioned in the original problem. When light strikes the grating, it bends around these lines. This bending is called diffraction.
  • The spacing between the lines in a grating determines how much the light will spread.
  • For a grating with 3850 lines/cm, the spacing is calculated as \(d = \frac{1}{38500} \, \text{m}\). This small spacing allows only certain wavelengths to constructively interfere, forming bright fringes at specific angles.
The angle at which these bright fringes appear can be used to determine the wavelength of the light, using the diffraction equation: \(d \sin \theta = m \lambda\).
Wien's Displacement Law
Wien's displacement law is fundamental in understanding blackbody radiation. It describes the relationship between the temperature of a blackbody and the peak wavelength \(\lambda_{max}\) of the radiation it emits.
  • The formula for Wien's displacement law is \(\lambda_{max} T = b\), where \(b = 2.8977719 \times 10^{-3} \, \text{m} \cdot \text{K}\). This value is constant for all blackbodies.
  • Using this law, once you know the peak wavelength from the diffraction grating analysis, you can calculate the temperature (T) of the blackbody: \(T = \frac{b}{\lambda_{max}}\).
This relationship helps us understand that as a blackbody's temperature increases, the peak wavelength of its emitted radiation decreases, shifting towards the blue or violet part of the spectrum.
Stefan-Boltzmann Law
The Stefan-Boltzmann law explains how the total energy radiated per unit surface area of a blackbody is proportional to the fourth power of its temperature. The law is formulated as \(P = \sigma A T^4\), where:
  • \(P\) is the power radiated per unit area.
  • \(\sigma = 5.67 \times 10^{-8} \, \text{W/m}^2\cdot\text{K}^4\) is the Stefan-Boltzmann constant.
  • \(A\) is the surface area of the blackbody.
  • \(T\) is the absolute temperature of the surface.
This law can be used to find out how much energy the blackbody sphere emits per second by substituting the area and temperature calculated earlier. Therefore, to determine how long it takes to radiate a given amount of energy, one needs to find the energy output rate (or power) and relate it to the total energy requirement.
First-Order Diffraction
First-order diffraction refers to the initial and most prominent set of diffraction maxima observed after the central or zero-order maximum. When analyzing light with a diffraction grating, the diffraction order corresponds to the integer \(m\) in the diffraction equation \(d \sin \theta = m \lambda\).
  • In the first-order diffraction, \(m = 1\), which makes it easier to pinpoint the angle \(\theta\) associated with the first bright band next to the central peak.
  • The angle at which this occurs provides essential information about the wavelength of light when observed through the grating.
  • In practical applications, including our example problem, identifying the angle of first-order diffraction can lead to determining properties like a blackbody's temperature through its emitted wavelength.
Being able to isolate the first-order diffraction help researchers to focus only on the most significant portion of the spectrum for further analysis.

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Most popular questions from this chapter

A beam of protons and a beam of alpha particles (of mass \(6.64 \times 10^{-27} \mathrm{kg}\) and charge \(+2 e )\) are accelerated from rest through the same potential difference and pass through identical circular holes in a very thin, opaque film. When viewed far from the hole, the diffracted proton beam forms its first dark ring at \(15^{\circ}\) with respect to its original direction. When viewed similarly, at what angle will the alpha particle form its first dark ring?

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(a) Approximately how fast should an electron move so it has a wavelength that makes it useful to measure the distance between adjacent atoms in typical crystals (about 0.10 \(\mathrm{nm} ) ?\) (b) What is the kinetic energy of the electron in part (a)? (c) What would be the energy of a photon of the same wavelength as the electron in part (b)? (d) Which would make a more effective probe of small- scale structures: electrons or photons? Why?

A \(10.0-\mathrm{g}\) marble is gently placed on a horizontal tabletop that is 1.75 \(\mathrm{m}\) wide. (a) What is the maximum uncertainty in the horizontal position of the marble? (b) According to the Heisenberg uncertainty principle, what is the minimum uncertainty in the horizontal velocity of the marble? (c) In light of your answer to part (b), what is the longest time the marble could remain on the table? Compare this time to the age of the universe, which is approximately 14 billion years. (Hint: Can you know that the horizontal velocity of the marble is exactly zero?)
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