Discussion
A plasma runs at about \( 7000 \ \mathrm{K} \) against a flame's \( 2500 \ \mathrm{K} \). The excited fraction for a two electron volt transition rises by roughly
Answer: C
Comparing the Boltzmann factors at the two temperatures:
\[ \begin{aligned} \frac{e^{-2.1/kT_{p}}}{e^{-2.1/kT_{f}}} \\ &\approx \frac{0.032}{6\times10^{-5}} \end{aligned} \]
which is a factor of several hundred. That exponential dependence on temperature is why emission techniques gain so enormously from a hotter source, and why absorption, which relies on the ground state, gains almost nothing.
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