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 .

0.00073

Explanation:

= Fraction of atoms of the element in the excited state

= Fraction of states

T = Temperature = 4471 K

h = Planck's constant =

= Boltzmann constant =

= Wavelength = 407.3 nm

We have the relation

Change in energy is given by

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

a. vertically

explanation:

i tried it and took the test

if im wrong then we have different tests