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The wavelength corresponding to maximum spectral radiance of a black body \(A\) is \(\lambda_{A}=5000 \AA\). Consider another black body \(B\), whose surface area is twice that of \(A\) and total radiant energy by \(B\) is 16 times that emitted by \(A\). The wavelength corresponding to maximum spectrum radiance for \(B\) will be (A) \(5000(2)^{3 / 4} \AA\) (B) \(2500 \AA\) (C) \(10,000 \AA\) (D) \(5000(2)^{-3 / 4} \AA\)

Short Answer

Expert verified
The wavelength corresponding to maximum spectral radiance for black body B is \(5000(2)^{-3 / 4} \AA\), so the correct answer is (D) \(5000(2)^{-3 / 4} \AA\).

Step by step solution

01

Formulate the relation between energy, surface area and temperature for a black body

By Stefan-Boltzmann law, the total energy emitted by a black body is given by \(E=\sigma A T^{4}\) where \(A\) is the surface area, \(T\) is the temperature, and \(\sigma\) is the Stefan-Boltzmann constant.
02

Find the relation between the temperatures of body A and B

According to the problem, the total radiant energy by B is 16 times that emitted by A. So, we have \(\sigma \cdot 2A \cdot T_{B}^{4} =16 \sigma \cdot A \cdot T_{A}^{4}\). With \(\sigma\) and \(A\) cancelled out, it becomes \(T_{B}=\sqrt[4]{2} T_{A}\), which means the temperature of B is \(\sqrt[4]{2}\) times the temperature of A.
03

Apply Wien's displacement law

Wien's displacement law states that the wavelength of maximum spectral radiance is inversely proportional to the temperature, hence \(\lambda_{B} T_{B}=\lambda_{A} T_{A}\). Substituting the value of \(T_{B}\) from step 2, we get \(\lambda_{B} = \frac{\lambda_{A}}{\sqrt[4]{2}} = 5000(2)^{-\frac{3}{4}} \AA\).

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Key Concepts

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

Stefan-Boltzmann Law
Understanding the Stefan-Boltzmann law is crucial when studying the properties of black body radiation. In simple terms, this law explains how the total energy radiated per unit surface area of a black body is directly proportional to the fourth power of the black body's thermodynamic temperature. Mathematically, it's expressed as:
\[ E = \sigma A T^{4} \]
In this formula, \( E \) is the total energy radiated per unit time, \( \sigma \) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8}\) W/m\(^2\)K\(^4\)), \( A \) is the surface area of the body, and \( T \) is the absolute temperature in Kelvin.
For example, if two black bodies are being compared and one radiates more energy than the other, the law indicates that this body has a higher temperature. This fundamental concept is vital for students preparing for competitive exams such as JEE MAIN Physics, where problems often revolve around the applications of this law.
Wien's Displacement Law
Wien's displacement law serves as a powerful tool in understanding black body radiation. This law establishes a relationship between the temperature of a black body and the peak wavelength of the emitted spectrum. It indicates that the black body radiation curve for different temperatures peaks at wavelengths inversely proportional to the temperature. The law is given by:
\[ \lambda_{max} = \frac{b}{T} \]
Where \( \lambda_{max} \) is the peak emission wavelength, \( T \) is the absolute temperature, and \( b \) is Wien's displacement constant (approximately \(2.897 \times 10^{-3}\) mK).
This law explains why, for instance, a hotter object may emit radiation with a shorter peak wavelength, often placing it in the visible spectrum, therefore appearing bluish, while cooler objects emit longer wavelengths, suggesting redder hues. Again, problems involving Wien's law frequently appear in JEE MAIN Physics exams, where students must analyze changes in temperature and their effects on peak wavelengths.
Spectral Radiance
Spectral radiance is a term used to describe the amount of energy a black body emits at a specific wavelength. It varies with both the temperature of the black body and the wavelength of the emission. The spectral radiance peak is the wavelength at which the radiation emitted by a black body is maximum and this peak shifts in response to changes in the body's temperature, as indicated by Wien's displacement law.
In practice, if we have two black bodies of different temperatures, even if they are made of the same materials, their spectral radiance curves will not be the same; the hotter body will reach its peak at a shorter wavelength compared to the cooler body. Students tackling JEE MAIN Physics problems will need to recognize how spectral radiance changes with temperature to predict radiation behaviors of various materials.
JEE MAIN Physics
The Joint Entrance Examination (JEE) MAIN for Physics is known to test students' knowledge and understanding of various physics concepts, including those related to black body radiation. The sections dealing with thermodynamics and modern physics often include questions about the Stefan-Boltzmann law, Wien's displacement law, and spectral radiance among others.
Preparing for these topics requires a depth of understanding and the ability to apply the concepts to solve complex problems. For example, the exercise mentioned earlier combines principles from both the Stefan-Boltzmann law and Wien's displacement law to determine the peak wavelength of emission from two black bodies under different conditions. Problems like these assess a student's ability to synthesize information and apply mathematical reasoning to physical concepts—a skill that is core to succeeding in the JEE MAIN Physics examination.

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Most popular questions from this chapter

The energy spectrum of a black body exhibits maximum around a wavelength \(\lambda_{0}\). The temperature of the black body is now changed such that the energy is maximum around a wavelength \(\frac{3 \lambda_{0}}{4} .\) The power radiated by the black body will now increase by a factor of (A) \(\frac{256}{81}\) (B) \(\frac{64}{27}\) (C) \(\frac{16}{9}\) (D) \(\frac{4}{3}\)

In hydrogen like atom an electron is orbiting in an orbit having quantum number \(n\). Its frequency of revolution is found to be \(13.2 \times 10^{15} \mathrm{~Hz}\). Energy required to free the electron from the atom from the above orbit is \(54.4 \mathrm{eV}\). In time 7 nano second the electron jumps back to orbit having quantum number \(\frac{n}{2} . \tau\) be the average torque acted on the electron during the above process, then find \(\tau \times 10^{27}\) in \(\mathrm{Nm}\). (given : \(\frac{h}{\pi}=\) \(2.1 \times 10^{-34} \mathrm{~J}-\mathrm{s}\), frequency of revolution of electron in the ground state of H-atom \(v_{0}=6.6 \times 10^{15} \mathrm{~Hz}\) and ionization energy of \(\mathrm{H}\)-atom, \(E_{0}=13.6 \mathrm{eV}\) ).

If the maximum kinetic energy of emitted photo electrons from a metal surface of work function \(2.5 \mathrm{eV}\), is \(1.7 \mathrm{eV}\). If wavelength of incident radiation is halved, then stopping potential will be (A) \(2.5 \mathrm{~V}\) (B) \(6.7 \mathrm{~V}\) (C) \(5 \mathrm{~V}\) (D) \(1.1 \mathrm{~V}\)

If doubly ionized lithium atom is hydrogen like with atomic number 3 , the wavelength of radiation required to excite the electron in \(\mathrm{Li}^{\mathrm{H}}\) from the first to the third Bohr orbit and the number of different spectral lines observed in the emission spectrum of the above excited system are (A) \(296 \AA, 6\) (B) \(114 \AA, 3\) (C) \(1026 \AA, 6\) (D) \(8208 \AA, 3\)

Due to annihilation of electron-positron of same kinetic energy \(0.95 \mathrm{MeV}\), a photon is produced which can also be produced by a photo- electron of energy \(E\), the possible value(s) of \(E\) is/are (mass of electron \(=9.1 \times 10^{-31} \mathrm{~kg}, e=1.6 \times 10^{-19}\) Coulomb) (A) \(1.02 \mathrm{MeV}\) (B) \(2.42 \mathrm{MeV}\) (C) \(4.03 \mathrm{MeV}\) (D) \(2.93 \mathrm{MeV}\)

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