Question

Consider an element with energy levels E 0 and E ∗ and degeneracies of those energy levels g 0 and g ∗ , respectively. Determine the fraction of atoms of the element in the excited state ( N ∗ / N 0 ) at 4471 K if the wavelength difference of the two states is 407.3 nm, and g 0 = 2 and g ∗ = 4 .

Answer 1

Answer:

0.00073

Explanation:

$\dfrac{N^*}{N_0}$ = Fraction of atoms of the element in the excited state

$\dfrac{g^*}{g_0}=\dfrac{4}{2}$ = Fraction of states

T = Temperature = 4471 K

h = Planck's constant = $6.626\times 10^{-34}\ m^2kg/s$

$k$ = Boltzmann constant = $1.38\times 10^{-23}\ J/K$

$\lambda$ = Wavelength = 407.3 nm

We have the relation

$\dfrac{N^*}{N_0}=\dfrac{g^*}{g_0}e^{-\dfrac{\Delta E}{kt}}$

Change in energy is given by

$\Delta E=\dfrac{hc}{\lambda}\\\Rightarrow \Delta E=\dfrac{6.626\times 10^{-34}\times 3\times 10^{8}}{407.3\times 10^{-9}}$

$\dfrac{N^*}{N_0}=\dfrac{4}{2}e^{-\dfrac{\dfrac{6.626\times 10^{-34}\times 3\times 10^{8}}{407.3\times 10^{-9}}}{1.38\times 10^{-23}\times 4471}}\\\Rightarrow \dfrac{N^*}{N_0}=0.00073$

The fraction of atoms of the element in the excited state is 0.00073