FLAME TEST

- range of wavelengths of electromagnetic radiation ATOMIC SPECTRA - range of wavelengths of electromagnetic radiation - wavelengths of visible light are separated when a beam of white light passes through a prism

SPECTROSCOPE

WAVELENGTH (λ) – distance between two adjacent crests or two adjacent troughs FREQUENCY (υ) – number of cycles or waves per second SPEED OF LIGHT (c) – 3 x 10 8 m/s c = λ υ

ENERGY LEVELS of ELECTRONS in HYDROGEN 1 -2.178 x 10 -18 J 2 -5.445 x 10 -19 J 3 -2.420 x 10 -19 J 4 -1.361 x 10 -19 J 5 -8.712 x 10 -20 J 6 -6.050 x 10 -20 J

To solve for the wavelength of light emitted by a change in the energy level of the electron, use the formula: λ = hc ΔE where h = 6.626 x 10 -34 J/s (Planck’s constant) c = 3.00 x 10 8 m/s (speed of light) ΔE = change in energy λ = wavelength of photon emitted in nanometers (nm)

Find the wavelength of a photon emitted when an electron jumps from the n = 3 energy level down to the n = 2 energy level. Where is this photon in the electromagnetic spectrum? λ = hc = (6.626 x 10 -34 J/s) (3.00 x 10 8 m/s) ΔE (-2.420 x 10 -19 J) - (-5.445 x 10 -19 J) = 6.571 x 10 -7 m or 657.1 nm