Question 20.4: Thermal Excitation of Atomic Energy Levels As discussed in S...
Thermal Excitation of Atomic Energy Levels
As discussed in Section 20.3, atoms can occupy only certain discrete energy levels. Consider a gas at a temperature of 2 500 K whose atoms can occupy only two energy levels separated by 1.50 eV, where 1 eV (electron volt) is an energy unit equal to 1.60 \times 10^{-19} J. Determine the ratio of the number of atoms in the higher energy level to the number in the lower energy level.
Conceptualize In your mental representation of this example, remember that only two possible states are allowed for the system of the atom. Figure 20.9 helps you visualize the two states on an energy-level diagram. In this case, the atom has two possible energies, E_1 and E_2, where E_1<E_2.
Categorize We categorize this example as one in which we focus on particles in a two-state quantized system. We will apply the Boltzmann distribution law to this system.
Analyze Set up the ratio of the number density of atoms in the higher energy level to the number density in the lower energy level and use Equation 20.41 to express each number:
n_V(E)=n_0 e^{-E / k_{B} T} (20.41)
(1) \frac{n_V\left(E_2\right)}{n_V\left(E_1\right)}=\frac{n_0 e^{-E_2 / k_{B} T}}{n_0 e^{-E_1 / k_{B} T}}=e^{-\left(E_2-E_1\right) / k_{B} T}
Evaluate k_{B} T in the exponent:
k_{B} T=\left(1.38 \times 10^{-23} J /K\right)(2 500 K)\left(\frac{1 eV}{1.60 \times 10^{-19} J}\right)=0.216 eVSubstitute this value into Equation (1):
\frac{n_V\left(E_2\right)}{n_V\left(E_1\right)}=e^{-1.50 eV/ 0.216 eV}=e^{-6.96}=9.52 \times 10^{-4}Finalize This result indicates that at T = 2 500 K, only a small fraction of the atoms are in the higher energy level. In fact, for every atom in the higher energy level, there are about 1 000 atoms in the lower level. The number of atoms in the higher level increases at even higher temperatures, but the distribution law specifies that at equilibrium there are always more atoms in the lower level than in the higher level.
