Spectrum of the hydrogen atom

Emission spectrum of the hydrogen atom, with three spectrum series:  Lyman (UV); Balmer (optical); and Paschen (IR)
The hydrogen atom offers the simplest example of emission and absorption because it has only one proton in the nucleus and one electron orbiting around it. Some of the unusual features of the quantum theory were accepted remarkably quickly because of the success of Bohr’s model in matching the observed wavelengths of the line spectrum for hydrogen. Read the text aloud
Hydrogen gas emits spectral lines at ultraviolet (UV), visible, and infrared wavelengths. Atomic transitions to and from the n = 1 energy level correspond to wavelengths of light in the ultraviolet and are named the Lyman series. Atomic transitions between the n = 2 energy level and higher n levels correspond to wavelengths of light in the visible and near-UV and are named the Balmer series. Atomic transitions between n = 3 and higher quantum numbers correspond to photon energies in the infrared. The lines corresponding to the Lyman series have the highest energy, followed by the lines of the Balmer series. Read the text aloud Show Balmer series of hydrogen
A hydrogen atom has an electron at an energy level of −5.45×10−19 J. It suddenly makes a transition to the −2.18×10−18 J energy level.
(a) Did it emit or absorb a photon? (b) What is the energy of the photon?
(c) What is the photon’s wavelength?  
(d) What part of the electromagnetic spectrum does this wavelength correspond to?
Asked: (a) whether it emits or absorbs a photon; (b) energy of the photon;
(c) wavelength of the photon; (d) region of the electromagnetic spectrum
Given: initial energy level of −5.45×10−19 J, final energy level of −2.18×10−18 J
Relationships: Planck equation E = hf, speed of light equation c =
Solution: (a) The atom lost energy; therefore, it emitted a photon.
(b) The atom’s change in energy equals the photon’s energy: E photon =5.45× 10 19  J(2.18× 10 18  J)=1.64× 10 18  J (c) Substitute the speed of light equation into the Planck equation to get ΔE = hc/λ. Solving for λ gives λ= hc ΔE = (6.63× 10 34  J s)(3× 10 8  m/s) 5.45× 10 19  J(2.18× 10 18  J) =121.7 nm (d) This wavelength is in the ultraviolet on the figure on page 1278.
Answer: (a) It emitted the photon; (b) 1.64×10−18 J; (c) 121.7 nm; and
(d) ultraviolet portion of the EM spectrum.
Read the text aloud Show Sign for the change in energy
Which of the following electron transitions in a hydrogen atom would emit visible light?
  1. n = 1 → n = 2
  2. n = 3 → n = 1
  3. n = 4 → n = 2
  4. n = 5 → n = 3
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