What wavelength of light is emitted when an electron falls from the n=4 to the n=3 orbital of the Lithium +2 ion (the ionization energy is 11815 kJ/mol)?

Answer 1

First, we should convert ionization energy from kJ/mole to kJ/atom:

11815 kJ/mol x 1 mol/6.02x10²³ atoms = 1.96x10^-20 kJ/atom

The energy of the photon is proportional to frequency as given by E = hν where E = energy of photon, h = Planck's constant and ν = frequency.

E = hν

ν = E/h = 1.96x10^-17 J / 6.626x10^-34 Jsec (note: changed E to joules to be consistent with Planck's constant)

ν = 2.96x10¹⁶ sec^-1

This would be the frequency. To find wavelength, we use the following;

λ = c/ν

λ = wavelength

c = speed of light

ν = frequencey

λ = c/ν = 3x10⁸ msec^-1 / 2.96x10¹⁶ sec^-1

λ = 1x10^-8 m