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Chapter 18: Problem 53

Surface temperature of the sun as estimated is \(6032.25 \mathrm{~K}\). The wavelength at which sun radiates maximum energy, is: (Given: Wein's constant \(=0.2898 \mathrm{~cm}-\mathrm{K}\) (a) \(\lambda_{m}=5000 \AA\) (b) \(\lambda_{m}=4804.2 \AA\) (c) \(\lambda_{m}=3809.5 \AA\) (d) \(\lambda_{m}=2891.6 \AA\)

Short Answer

Expert verified
The wavelength is approximately 4803.8 脜, closest to option (b) 4804.2 脜.

Step by step solution

01

Understand Wien's Displacement Law

Wien's Displacement Law allows us to find the wavelength at which a black body emits the maximum amount of energy. It is given by the formula: \( \lambda_{m} T = b \), where \( \lambda_{m} \) is the wavelength in meters, \( T \) is the temperature in Kelvin, and \( b \) is Wien's constant \( 0.2898 \text{ cm-K} \) or \( 2.898 \times 10^{-3} \text{ m-K} \).
02

Substitute the Given Values

Substitute the given temperature of the sun \( T = 6032.25 \text{ K} \) and Wien's constant \( b = 2.898 \times 10^{-3} \text{ m-K} \) into the formula. This gives us: \( \lambda_{m} \times 6032.25 = 2.898 \times 10^{-3} \).
03

Calculate Wavelength in Meters

To find \( \lambda_{m} \), rearrange the equation to: \( \lambda_{m} = \frac{2.898 \times 10^{-3}}{6032.25} \). Perform the division to get \( \lambda_{m} \approx 4.8038 \times 10^{-7} \text{ m} \).
04

Convert Wavelength to Angstrom

Convert the wavelength from meters to Angstroms. Since \( 1 \text{ m} = 10^{10} \text{ 脜} \), \( \lambda_{m} \approx 4.8038 \times 10^{-7} \text{ m} = 4803.8 \text{ 脜} \).
05

Compare with Given Options

Compare the calculated wavelength \( 4803.8 \text{ 脜} \) with the given options. The closest value is \( \lambda_{m} = 4804.2 \text{ 脜} \), which corresponds to option (b).

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Key Concepts

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

Black Body Radiation
Black body radiation is a fascinating concept in physics that refers to the type of electromagnetic radiation emitted by a perfect black body. A black body is an idealized physical object that absorbs all incident radiation, regardless of frequency or angle. Because it absorbs perfectly, it also emits radiation perfectly.
  • A perfect black body emits radiation in a predictable pattern, determined by its temperature.
  • The emitted radiation covers a wide range of wavelengths.
  • As the temperature increases, the peak wavelength of the emitted radiation shifts to shorter wavelengths (more energetic).
This property of black body radiation helps scientists understand and predict how real-world objects, like stars, emit energy. Wien's Displacement Law, which we'll discuss next, plays a key role in this understanding by providing a relationship between the temperature of a black body and the wavelength at which it emits the maximum energy.
Sun's Surface Temperature
The surface temperature of the sun is an essential factor that influences the characteristics of the radiation it emits. Estimated to be around 6032.25 Kelvin, the sun's surface temperature plays a critical role in determining the spectrum of light it radiates. At such high temperatures:
  • The sun emits a great amount of energy across a wide range of wavelengths.
  • The dominant (or peak) wavelength of this emission gives the sun its characteristic yellow-white color.
  • Understanding this temperature allows astronomers to deduce various properties of the sun and other stars, such as age and composition.
Wien's Displacement Law, which we used earlier, gives us a convenient formula to link this temperature directly to the peak emission wavelength, illustrating how theoretical physics can be used to predict observable astronomical phenomena.
Wavelength Conversion
In the context of this exercise, we performed a wavelength conversion from meters to angstroms. Wavelength conversion is crucial for interpreting astronomical data because different measurements and scientific communities may use different units.Here's a quick guide on performing these conversions:
  • 1 meter is equivalent to 10 billion angstroms (1 m = \(10^{10}\) 脜).
  • Conversion involves multiplying the measurement in meters by \(10^{10}\) to switch to angstroms.
  • This practice is common in astronomy because angstroms are very convenient for expressing wavelengths of visible light.
Being adept at unit conversions like this helps ensure accurate communication and comparison of scientific data, making it easier for scientists around the world to share and build upon each other's work.
Spectral Distribution
Spectral distribution refers to the way that energy emitted by a black body is spread across different wavelengths. This distribution can be visualized through a spectrum, showing how much light is emitted at each wavelength. The key points to understand about spectral distribution include:
  • Temperature significantly affects the spectral distribution of a black body.
  • A hotter object peaks at shorter wavelengths, meaning it emits more high-energy, short-wavelength radiation.
  • The sun, at a surface temperature of about 6032.25 K, has its peak emission in the visible light region of the spectrum.
Wien's Displacement Law helps illustrate the peak, or most intense part, of this distribution. By knowing the peak wavelength and temperature, scientists can derive many useful insights about the physical properties and behaviors of stars and other planetary bodies.

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Most popular questions from this chapter

A body of length \(1 \mathrm{~m}\) having cross-sectional area \(0.75 \mathrm{~m}^{2}\) has thermal conductivity \(6000 \mathrm{~J} / \mathrm{s}\) then the temperature difference if \(\mathrm{K}=200 \mathrm{Jm}^{-1} \mathrm{~K}^{-1}\), will be : (a) \(20^{\circ} \mathrm{C}\) (b) \(40^{\circ} \mathrm{C}\) (c) \(80^{\circ} \mathrm{C}\) (d) \(100^{\circ} \mathrm{C}\)

In a composite rod, when two rods of different lengths and of the same area of cross-section, are joined end to end then if \(K\) is the effective coefficient of thermal conductivity \(\frac{l_{1}+l_{2}}{K}\) is equal to : (a) \(\frac{l_{1}}{K_{1}}-\frac{l_{2}}{K_{2}}\) (b) \(\frac{l_{1}}{K_{2}}=\frac{l_{2}}{K_{1}}\) (c) \(\frac{l_{1}}{K_{1}}+\frac{l_{2}}{K_{2}}\) (d) \(\frac{l_{1}}{K_{2}}+\frac{l_{2}}{K_{1}}\)

Though air is bad conductor yet a body kept in air losses heat quickly this is due to : (a) conduction (b) convection (c) radiation (d) none of these

The time in which a layer of ice of thickness \(h\) will grow on the surface of the pond of surface area \(A\), when the surrounding temperature falls to \(-T^{\circ} \mathrm{C}\) is : (Assume \(K=\) thermal conductivity of ice, \(\rho=\) density fo water, \(L=\) latent heat of fusion) (a) \(t=\frac{p L}{2 K T} h^{2}\) (b) \(t=\frac{\rho L}{K T} h^{2}\) (c) \(t:=\frac{\rho L h^{2}}{3 K T}\) (d) \(t=\frac{\rho L h^{2}}{4 K T}\)

\(1 \mathrm{~g}\) of ice at \(0^{\circ} \mathrm{C}\) is mixed with \(1 \mathrm{~g}\) of steam at \(100^{\circ} \mathrm{C}\). After thermal equilibrium is attained, the temperature of mixture will be : (a) \(1^{\circ} \mathrm{C}\) (b) \(50^{\circ} \mathrm{C}\) (c) \(81^{\circ} \mathrm{C}\) (d) \(100^{\circ} \mathrm{C}\)
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