Answer:
(a) N = 1.159x10⁻¹⁶³ atoms
(b) N = 6.9x10¹⁷ atoms
Explanation:
The Boltzmann factor is the following:
[tex] N_{E} = N_{i} e^{-\frac{E}{k_{B}T}} [/tex]
where [tex]N_{E}[/tex]: is the number of atoms in the energy state E, [tex]N_{i}[/tex]: is the number of atoms in the energy ground state, [tex]k_{B}[/tex]: is the Boltzmann constant and T: is the temperature
(a) The number of atoms in the first excited state at 0°C is:
[tex] N_{10.2 eV} = 2.70 \cdot 10^{25} \cdot e^{-\frac{10.2 eV\cdot \frac{1.602 \cdot 10^{-19} J}{1 eV}}{(1.38 \cdot 10^{-23} J/K)(273 K)}} = 1.159 \cdot 10^{-163} atoms [/tex]
The number of atoms in the first excited state at 0 °C is approximately zero.
(b) The number of atoms in the first excited state at 6.50x10³ °C is:
[tex] N_{10.2 eV} = 2.70 \cdot 10^{25} \cdot e^{-\frac{10.2 eV\cdot \frac{1.602 \cdot 10^{-19} J}{1 eV}}{(1.38 \cdot 10^{-23} J/K)((6.50 \cdot 10^{3} + 273) K)}} = 6.9 \cdot 10^{17} atoms [/tex]
The increasing of the temperature to 6.50x10³ °C results in an increase in the number of atoms in the first excited state to 6.9x10¹⁷ atoms.
I hope it helps you!
