Question

Metallic elements often show visible emission lines when their salts are vaporized in a Bunsen burner flame, and the distinctive colors produced are the basis for the visual flame tests for the presence of various elements. Lithium salts show a bright red flame (λ = 671 nm), sodium yellow-orange (589 nm), and potassium a faint red (764 nm). Calculate ∆E in electron-volts and inverse centimeters for the three elements.

Answer 1

For Lithium atom,

= 671 nm = 671 * 10^-9 m = 671 * 10^-7 cm

and E= hc/

Wher h = planck's constant = 6.626 * 10^-34 J.s

c= velosity of light = 3.0 *10⁸m/s

So, E= ( 6.626 * 10^-34 J.s * 3.0 *10⁸m/s)/ (671 * 10^-9 m) = 2.96 * 10^-19 J

1 J = 6.242 * 10¹⁸ eV

So, 2.96 * 10^-19 J = 6.242 * 10¹⁸ * 2.96 * 10^-19 = 1.848 eV

E =  1.848 eV

wavenumber = 1/ = 1/ ( 671 * 10^-7 m) = 1.49 * 10⁴ cm^-1

For Sodium

= 589 nm = 589 * 10^-9 m = 589 * 10^-7 cm

E= ( 6.626 * 10^-34 J.s * 3.0 *10⁸m/s)/ (589 * 10^-9 m) = 3.37 * 10^-19 J = 3.37 * 10^-19 J * 6.242 * 10¹⁸ = 2.05 eV

wavenumber = 1/ = 1/ ( 589 * 10^-7 m) = 1.7 * 10⁴ cm^-1

For Potassium

= 764 nm = 764 * 10^-9 m = 764 * 10^-7 cm

E= ( 6.626 * 10^-34 J.s * 3.0 *10⁸m/s)/ (764 * 10^-9 m) = 2.60 * 10^-19 J = 2.60* 10^-19 J * 6.242 * 10¹⁸ = 1.62 eV

wavenumber = 1/ = 1/ ( 764 * 10^-7 m) = 1.31 * 10⁴ cm^-1