Which target will emit more overall energy (i.e., energy over all wavelengths)?

The target that emits more overall energy, or energy across all wavelengths, is determined by the principles of blackbody radiation, specifically described by Planck's law and Stefan-Boltzmann law. According to these principles, an increase in temperature leads to a significant increase in the amount of energy emitted.

The Stefan-Boltzmann law states that the total energy radiated per unit surface area of a black body is proportional to the fourth power of its absolute temperature. Therefore, as the temperature increases, the energy output increases exponentially.

In this scenario, the 1200 K exhaust plume will emit significantly more energy than the 900 K exhaust plume. This is because, when applying the Stefan-Boltzmann law, the emission from the 1200 K plume would be:

E_1200K = σ * (1200^4)

E_900K = σ * (900^4)

Where σ is the Stefan-Boltzmann constant. When you compare these two, the energy emitted by the 1200 K plume vastly surpasses that of the 900 K plume due to the fourth power relationship. This is why the 1200 K exhaust plume is the correct answer as it emits more overall energy.