The energy diﬀerence between the ﬁrst excited state of mercury and the ground state is 4.86 eV.
(a) If a sample of mercury vaporized in a ﬂame contains 10^20
atoms in thermal equilibrium at 1600K, calculate the number of atoms in the n=1 (ground) and n=2 (ﬁrst-excited) states. (Assume the Maxwell-Boltzmann distribution applies and that the n=1 and n=2 states have equal statistical weights.)
Maxwell Boltzmann Distribution
The Attempt at a Solution
I thought that since they have the same statistical weight, there must be 5*10^19 particles in each state. But I don't think it is the good answer since we use this number for another exercice and it doesn't yield the good answer.
I don't know how to figure out these number using Maxwell Boltzmann distribution.