Question

A hydrogen atom emits a photon of energy about 13.22 eV. What are the values of n for the initial and final states involved in this transition?
n_i =      
[5 points] 0 attempt(s) made (maximum allowed for credit = 5)

n_f =      
[5 points] 0 attempt(s) made (maximum allowed for credit = 5)

Answer 1

A hydrogen atom in the ground state absorbs a photon of wavelength 93.8 nm. The excited atom then emits a photon of wavelength 7459 nm. What energy level does the electron reach initially (n initial), and what energy level does the electron fall to after the emission (n final)?

So you can use the Bohr formula

En = -13.6*(1/n2^2 - 1/n1^2) eV where En is in electron volts.

Energy of a photon is Ep = 1.24/wl eV where wl is wavelength in microns

So 93.8 nm = 0.0938 um so Ep = 13.22 ev. Now atom radiates a photon wl' = 7.459 um loosing an energy E' of

E' = 1.24/7.459 = 0.166 eV

SO the atom has a net gain in energy of Ep - E' =13.053 eV

TO find th eenergy level, set Ep - E' = En

13.053 = -13.6 (1/n2^2 - 1/n1^2) and set n1 = 1 = ground state. then solve for n2


-13.055/13.6 = 1/n2^2 - 1 --> 1 - 13.055/13.6 = 1/n2^2

0.040193 = 1/n2^2 ---> n2 = sqrt(1/0.040193) = 4.98 ~ 5

So n final is 5