/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 A block body is at a temperature... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 6

A block body is at a temperature \(2880 \mathrm{~K}\). The energy radiation emitted by this object with wavelength between \(499 \mathrm{~nm}\) and \(500 \mathrm{~nm}\) is \(U_{1}\), between \(999 \mathrm{~mm}\) and \(1000 \mathrm{~nm}\) is \(U_{2}\), and between \(1499 \mathrm{~nm}\) and \(1500 \mathrm{~nm}\) is \(U_{3}\), then (Wien's constant \(b=2.88 \times 10^{6} \mathrm{~nm}-\mathrm{K}\) ) (A) \(U_{1}>U_{2}\) (B) \(U_{2}>U_{1}\) (C) \(U_{1}=0\) (D) \(U_{3}=0\)

Short Answer

Expert verified
(A) \(U_1 > U_2\).

Step by step solution

01

Write down Planck's Law formula

Planck's Law formula is given by: \[u(\lambda, T) = \frac{2 \pi h c^2}{\lambda^5} \frac{1}{e^{\frac{h c}{\lambda k_B T}} - 1}\] where \(u(\lambda, T)\) is the energy density emitted by the black body per unit wavelength interval, \(h\) is the Planck's constant (\(6.626 \times 10^{-34} \mathrm{Js}\)), \(c\) is the speed of light (\(2.998 \times 10^8 \mathrm{m/s}\)), \(\lambda\) is the wavelength, \(T\) is the temperature of the black body in Kelvin, and \(k_B\) is the Boltzmann constant (\(1.381 \times 10^{-23} \mathrm{J/K}\)).
02

Calculate the energy radiation for each wavelength range

To find the energy radiation in each wavelength range, we will integrate Planck's Law over the given wavelength intervals. For \(U_1\), we need to integrate from \(\lambda = 499 \mathrm{nm}\) to \(\lambda = 500 \mathrm{nm}\): \[U_1 = \int_{499}^{500} u(\lambda, 2880) d\lambda\] For \(U_2\), we need to integrate from \(\lambda = 999 \mathrm{nm}\) to \(\lambda = 1000 \mathrm{nm}\): \[U_2 = \int_{999}^{1000} u(\lambda, 2880) d\lambda\] For \(U_3\), we need to integrate from \(\lambda = 1499 \mathrm{nm}\) to \(\lambda = 1500 \mathrm{nm}\): \[U_3 = \int_{1499}^{1500} u(\lambda, 2880) d\lambda\]
03

Calculate the wavelength of maximum energy emission

Using Wien's Displacement Law, we can find the wavelength at which the black body emits the maximum amount of energy: \[ \lambda_{max} = \frac{b}{T}\] where \(b = 2.88 \times 10^6 \mathrm{nm \cdot K}\) is the Wien's constant and \(T = 2880 \mathrm{K}\) is the temperature. \[ \lambda_{max} = \frac{2.88 \times 10^6 \mathrm{nm \cdot K}}{2880 \mathrm{K}} = 1000 \mathrm{nm}\]
04

Compare the energy radiation values and select the correct option

Since the maximum energy emission occurs at \(\lambda_{max} = 1000 \mathrm{nm}\), the energy radiation values will decrease as we move away from this wavelength (before and after). This means that \(U_1 > U_2\) in the given wavelength ranges, leading to option (A) being correct. Answer: (A) \(U_1 > U_2\).

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