Question

Two energy levels of an atomic system are separated by energy corresponding to frequency 3.0 × 1014 Hz. Assume that all atoms are in one or the other of these two energy levels, compute the fraction of atoms in the upper energy level at temperature 400 K. Take kB = 1.38×10−23JK−1 and h = 6.6×10−34Js.

Answer 1

Answer on Question #65231, Physics / Optics

Two energy levels of an atomic system are separated by energy corresponding to frequency $3.0 \times 10^{14} \mathrm{~Hz}$. Assume that all atoms are in one or the other of these two energy levels, compute the fraction of atoms in the upper energy level at temperature $400 \mathrm{~K}$. Take $k = 1.38 \times 10^{-23} \mathrm{JK}^{-1}$ and $h = 6.6 \times 10^{-34} \mathrm{Js}$.

Find: $\frac{\mathrm{N}_2}{\mathrm{N}_1} - ?$

Given:

$\Delta \nu = 3.0 \times 10^{14} \mathrm{~Hz}$

T=400 K

$k = 1.38 \times 10^{-23} \mathrm{~J} \times \mathrm{K}^{-1}$

$h = 6.6 \times 10^{-34} \mathrm{~J} \times \mathrm{s}$

Solution:

Boltzmann factor:

$\frac{\mathrm{N}_2}{\mathrm{N}_1} = \exp \frac{-\mathrm{E}_1 - \mathrm{E}_2}{\mathrm{kT}}$ (1), where level 2 ($\mathrm{N}_2$) is higher than level 1 ($\mathrm{N}_1$)

Energy difference:

$\mathrm{E}_1 - \mathrm{E}_2 = h \Delta \nu$ (2)

(2) in (1): $\frac{\mathrm{N}_2}{\mathrm{N}_1} = \exp^{-\frac{h \Delta \nu}{kT}}$ (3)

Of (3) $\Rightarrow \frac{\mathrm{N}_2}{\mathrm{N}_1} = \exp^{-\frac{6.6 \times 10^{-34} \times 3.0 \times 10^{14}}{1.38 \times 10^{-23} \times 400}} = \exp^{-35.9} \approx 0$