A small particle of interplanetary material is heated by friction from a temperature of 400 K to 4000 K as it falls into the atmosphere of Earth and produces a meteor or a shooting star in the sky. If this object behaves like a perfect blackbody over this short time, how will its emitted radiation change as it is heated?
A) The intensity of the radiation will rise by a factor of 100, while the peak wavelength of emitted light will become shorter by a factor of 100, moving from the infrared to the ultraviolet.
B) The intensity of the radiation will rise by a factor of 10,000, while its peak wavelength will become longer by a factor of 10, moving from the visible to the infrared of heat radiation.
C) The emitted intensity of the radiation will rise by a factor of 10,000, while its peak wavelength will become shorter by a factor of 10, from infrared to red visible light.
D) The intensity of the radiation will rise by a factor of 10, while its peak wavelength will become shorter by a factor of 10, moving from the infrared to red visible light.

Question

Quizplus · Accepted Answer

The intensity of radiation emitted by a blackbody is given by the Stefan-Boltzmann law, which states that the total energy radiated per unit surface area is directly proportional to the fourth power of the blackbody's temperature (E ∝ T^4). Therefore, when the temperature increases from 400 K to 4000 K (a factor of 10), the intensity of radiation increases by a factor of 10^4 or 10,000. According to Wien's displacement law, the wavelength at which the radiation is strongest is inversely proportional to the temperature (λ_max ∝ 1/T). Thus, as the temperature increases by a factor of 10, the peak wavelength becomes shorter by the same factor, shifting from infrared towards the visible spectrum, specifically towards red visible light.

A Small Particle of Interplanetary Material Is Heated by Friction