Question

Two spherical black-bodies A and B, having radii ${r_A}$ and ${r_B}$ , where ${r_B} = 2{r_A}$ emit radiations with peak intensities at wavelengths $400\,nm$ and $800\,nm$ respectively. If their temperature are ${T_A}$ and ${T_B}$ respectively in Kelvin scale, their emissive powers are ${E_A}$ and ${E_B}$ and energies emitted per second are ${P_A}$ and ${P_B}$ then:
(A) $\dfrac{{{T_A}}}{{{T_B}}} = 2$
(B) $\dfrac{{{P_A}}}{{{P_B}}} = 4$
(C) $\dfrac{{{E_A}}}{{{E_B}}} = 8$
(D) $\dfrac{{{E_A}}}{{{E_B}}} = 4$

Answer 1

Hint : In order to solve the problem we should first understand about the Thermal Radiation. The electromagnetic radiation produced by the thermal motion of particles in matter is known as thermal radiation. Thermal radiation is emitted by all materials with a temperature greater than absolute zero. Charge-acceleration or dipole oscillation occurs as a result of particle motion, resulting in electromagnetic radiation.

Complete Step By Step Answer:
Thermal characteristics are those of a material that are related to its heat conductivity. In other words, these are the qualities that a material exhibits when heat is applied to it. Thermal properties are part of the larger issue of material physical properties.
The black-body radiation curve for different temperatures will peak at distinct wavelengths that are inversely proportional to the temperature, according to Wien's displacement law. The Planck radiation law, which specifies the spectrum brightness of black-body radiation as a function of wavelength at any given temperature, causes that peak to shift. Wilhelm Wien discovered it several years before Max Planck produced the more general equation, and it describes the full change of the spectrum of black-body radiation toward shorter wavelengths as temperature rises.
The spectrum brightness of black-body radiation per unit wavelength, according to Wien's displacement law, peaks at the wavelength ${\lambda _{peak}}$ given by:
${\lambda _{peak}} = \dfrac{b}{T}$
T stands for absolute temperature. The Wien's displacement constant, b, is a proportionality constant equal to $\approx 2898\,\mu m \cdot K.$
Given:
Wavelength of black-body A is: ${\lambda _A} = 400\,nm$
Wavelength of black-body B is: ${\lambda _B} = \,800\,nm$
${r_B} = 2{r_A}$ .
According to Wien’s law: ${\lambda _{peak}} = \dfrac{b}{T}$
$\therefore {\lambda _A} = \dfrac{b}{{{T_A}}}$
$\therefore {\lambda _B} = \dfrac{b}{{{T_B}}}$
$\Rightarrow \dfrac{{{\lambda _A}}}{{{\lambda _B}}} = \dfrac{{\dfrac{b}{{{T_A}}}}}{{\dfrac{b}{{{T_B}}}}}$
$\Rightarrow \dfrac{{{\lambda _A}}}{{{\lambda _B}}} = \dfrac{{\dfrac{1}{{{T_A}}}}}{{\dfrac{1}{{{T_B}}}}}$
$\Rightarrow \dfrac{{{\lambda _A}}}{{{\lambda _B}}} = \dfrac{{{T_B}}}{{{T_A}}}$
$\Rightarrow \dfrac{{{T_A}}}{{{T_B}}} = \dfrac{{{\lambda _A}}}{{{\lambda _B}}} = \dfrac{{400}}{{800}} = \dfrac{1}{2}$
$\Rightarrow \dfrac{{{T_A}}}{{{T_B}}} = 2$
$Power = \sigma \times A \times {T^4}$
Where, $\sigma = \,5.670 \times {10^{ - 8\,}}W.{m^{ - 2}} \cdot {K^{ - 4}}$ is the Stefan-Boltzmann constant.
$T =$ Absolute temperature
$A =$ Surface area of object
$\dfrac{{{P_A}}}{{{P_B}}} = \dfrac{{\sigma \times {A_A} \times {{({T_A})}^4}}}{{\sigma \times {A_B} \times {{({T_B})}^4}}} = \dfrac{{4\pi {{({r_A})}^2}}}{{4\pi {{({r_B})}^2}}} \times {\left( {\dfrac{{{T_A}}}{{{T_B}}}} \right)^4}$
$\dfrac{{{P_A}}}{{{P_B}}} = {\left( {\dfrac{1}{2}} \right)^2} \times {2^4} = {2^2} = 4$
$\dfrac{{{P_A}}}{{{P_B}}} = 4$
So, option (A) and (B) are correct.

Note :
Blackbody radiation occurs when a radiation object has the physical features of a black body in thermodynamic equilibrium. The spectrum of blackbody radiation is described by Planck's law, which is purely dependent on the temperature of the object. The most likely frequency of the emitted radiation is determined by Wien's displacement equation, and the radiant intensity is determined by the Stefan–Boltzmann law.