Final answer:
The thermal radiation emitted by a blackbody when its temperature falls from TK to T/4K is E/256, as per the Stefan-Boltzmann law, which indicates that the energy emitted is proportional to the fourth power of its temperature. The correct answer is option (4).
Step-by-step explanation:
The student's question pertains to the thermal radiation emitted by a blackbody in relation to its temperature, as described by the Stefan-Boltzmann law.
This law states that the energy flux (F) from a blackbody is proportional to the fourth power of its absolute temperature (T), which can be mathematically expressed as F = σeT4, where σ represents the Stefan-Boltzmann constant (5.67 x 10-8 J/s · m2 · K4), e is the emissivity, and T is the temperature in kelvin.
σ is a constant, and since a blackbody has an emissivity (e) of 1, the equation simplifies to
F = σT4.
Therefore, if the black body's temperature falls from TK to T/4K, the thermal radiation emitted in W/m2 can be calculated by plugging these values into the equation and comparing the two resulting fluxes.
With the initial temperature being TK, the initial energy flux is
E = σ(TK)4.
After the temperature falls to T/4K, the new energy flux will be E' = σ(T/4K)4 = σT4/256.
Hence, the thermal radiation emitted at the lower temperature will be E/256, corresponding to choice (4).