Line Spectra

1. Line Spectra When the particles in the solid, liquid, or gas accelerate, they will produce EM waves. Electron orbit to orbit transitions in atoms (gasses) Applicable to the study of stars (gaseous objects)

2. Line Spectra Emission Spectrum In accordance with the Energy Minimum Principle, the electron will then “jump” to a lower energy state. In doing so, it will give up a specific amount of energy through the emission of a photon. E1 E3 – E2 = h f32 E2 E3

3. The electron in a hydrogen atom is in the second excited state. What are the energies of the photons that could be emitted as the electron moves to the ground state? The second excited state corresponds to n = 3. Also, recall that for hydrogen, En = -13.6 eV / n2 The electron can drop directly from the second excited state (n=3) to the ground state (n=1) emitting a photon. It could also drop from the second excited state (n=3) to the first excited state (n=2) emitting a photon, and then drop from the first excited state (n=2) to the ground state (n=1) emitting another photon. Therefore, the energy change for the electron in each case is ΔE3 to 1 = -13.6 eV (1/12 – 1/32 ) = -12.08 eV ΔE3 to 2 = -13.6 eV (1/22 – 1/32 ) = -1.889 eV ΔE2 to 1 = -13.6 eV (1/12 – 1/22 ) = -10.2 eV Each energy loss produces a photon with the equivalent amount of energy, so the photon energies are 12.08 eV, 1.89 eV and 10.2 eV.

4. What is the frequency of each photon emitted in the previous calculation? The three photon energies are 12.08 eV, 1.89 eV and 10.2 eV. Converting each to Joules gives the following energies E31 = 12.08 eV = (12.08 eV) (1.6 x 10-19 J / 1 eV) = 1.94 x 10-18 J E32 = 1.89 eV = (1.889 eV) (1.6 x 10-19 J / 1 eV) = 3.022 x 10-19 J E21 = 10.2 eV = (10.2 eV) (1.6 x 10-19 J / 1 eV) = 1.632 x 10-18 J Then, since E = hf E31 = hf = (6.63 x 10-34 J-sec) f → f31 = (1.93 x 10-18 J) / (6.63 x 10-34 J-sec) = 0.29 x 1016 Hz = 2.9 x 1015 Hz. Similarly, f32 = (3.022 x 10-19 J) / (6.63 x 10-34 J-sec) = 0.46 x 1015 Hz = 4.56 x 1014 Hz. f21 = (1.632 x 10-18 J) / (6.63 x 10-34 J-sec) = 0.246 x 1016 Hz = 2.46 x 1015 Hz.

5. What is the wavelength of each photon emitted in the previous calculation? The three photon frequencies are f31 = 2.9 x 1015 Hz f32 = 4.56 x 1014 Hz f21 = 2.4 x 1015 Hz Since λ = c/f, and using the fact that 1 Hz = 1 sec-1 λ31 = (3 x 108 m/sec)/(2.9 x 1015 sec-1 ) = 1.034 x 10-7 m = 103.4 nm (Ly β line) λ32 = (3 x 108 m/sec)/(4.56 x 1014 sec-1 ) = .657 x 10-6 m = 6.57 x 10-7 m = 657 nm (Hα line) λ21 = (3 x 108 m/sec)/(2.46 x 1015 sec-1 ) = 1.22 x 10-7 m = 122 nm (Ly α line) Compare these results with the charts in the book or the overheads