Atomic spectra are a result of energy level diagrams - quantum theory

Absorption vs. Emission spectra

Atomic Line Width Although theoretically, the line width of an atomic line should be very narrow nm (it’s quantized!) But… Many environmental effects cause these lines to be broadened nm 1.Uncertainty Principle 2.Collisions with other atoms in gas phase 3.Doppler Effect 4.Zeeman Effect The width of the line affects the LOD. Although theoretically, the line width of an atomic line should be very narrow nm (it’s quantized!) But… Many environmental effects cause these lines to be broadened nm 1.Uncertainty Principle 2.Collisions with other atoms in gas phase 3.Doppler Effect 4.Zeeman Effect The width of the line affects the LOD.

Doppler and Zeeman Doppler shift - frequency shifts due to movement of atom with respect to the detector Zeeman effect - splitting of electronic states due to external EM field Doppler shift - frequency shifts due to movement of atom with respect to the detector Zeeman effect - splitting of electronic states due to external EM field

Hollow cathode lamp Cathode made of specified metal (solid) - atoms are sputtered off and excited. Bulb contains low pressure (1-5 Torr) inert gas. Cathode made of specified metal (solid) - atoms are sputtered off and excited. Bulb contains low pressure (1-5 Torr) inert gas.

Line width and LOD The line width from your HCL must be more narrow than the line width of your sample atoms.

Boltzmann Equation We use heat (flame, furnace) to atomize our samples. In emission spectrometry, this is also our means of excitation. How does the T of the flame affect the intensity of the emission? N j =P j e (-Ej/kT) N 0 P 0 N j = # atoms in excited state, N 0 = # in ground state P j = # quantum states in excited state, P 0 = # in ground state. i.e. for p orbitals, P = 6; for s orbitals, P = 2 E j = energy difference between ground and excited states (J) k = Boltzmann constant = 1.28 x J/K T = Temperature (K) We use heat (flame, furnace) to atomize our samples. In emission spectrometry, this is also our means of excitation. How does the T of the flame affect the intensity of the emission? N j =P j e (-Ej/kT) N 0 P 0 N j = # atoms in excited state, N 0 = # in ground state P j = # quantum states in excited state, P 0 = # in ground state. i.e. for p orbitals, P = 6; for s orbitals, P = 2 E j = energy difference between ground and excited states (J) k = Boltzmann constant = 1.28 x J/K T = Temperature (K)

Element to be measured must be… Vaporized –Nebulizer (solution) –Electrothermal vaporization, ETV (solution) Can use microliters of solution instead of several mL –Hydride Generation (only certain elements who form easily vaporized hydrides) –Cold vapor (mercury) –Direct insertion (solid) –Laser ablation (solid) Vaporized –Nebulizer (solution) –Electrothermal vaporization, ETV (solution) Can use microliters of solution instead of several mL –Hydride Generation (only certain elements who form easily vaporized hydrides) –Cold vapor (mercury) –Direct insertion (solid) –Laser ablation (solid)

After it is vaporized it must be… Atomized/Excited - must be in free atom state to excite the valence electrons –Hot flame ( C) –ETV ( C) –Plasmas - the fourth state of matter Inductively coupled plasma, ICP ( C) Other plasmas - microwave-induced, DC current Electric spark - 40,000 (?) - used in metallurgy Atomized/Excited - must be in free atom state to excite the valence electrons –Hot flame ( C) –ETV ( C) –Plasmas - the fourth state of matter Inductively coupled plasma, ICP ( C) Other plasmas - microwave-induced, DC current Electric spark - 40,000 (?) - used in metallurgy

Continuous vs. Discrete Flame AAS results in continuous signal. ETV AAS results in discete signal. Flame AAS results in continuous signal. ETV AAS results in discete signal.

Atomic spectroscopy is selective. You know what you are looking for and you measure that particular wavelength.