Light of three different wavelengths—325 nm, 455 nm, and 632 nm—is shined on a metal surface. The observations for each wavelength, labeled A, B, and C, is as follows:
Observation A: No photoelectrons were observed.
Observation B: Photoelectrons with a kinetic energy of 155 kJ/mol were observed
Observation C: Photoelectrons with a kinetic energy of 51 kJ/mol were observed
Which observation corresponds to which wavelength of light?

That is, the energy (hf) of the incident photon is used to eject an electron from the surface of the metal ( the work function, Φ), and anything left over goes into the kinetic energy (KE) of the electron.

1. Calculate the relative energies of the three wavelengths

E = hf

fλ = c or f = c/λ. So,

E = (hc)/λ

The energy is inversely proportional to the wavelength. Thus, the order of increasing energies is

632 nm < 455 nm < 325 nm

2. Assign the radiation to the observations.

Solve the photoelectric equation for KE.

E = Φ + KE

KE = E - Φ

If E < Φ, the light does not have enough energy to eject a photoelectron.

Once E > φ, the greater the value of E, the greater the value of KE.

The order of energies is

A. No photoelectrons < C. KE = 51 kJ·mol⁻¹ < B. KE = 155 kJ·mol⁻¹

632 nm < 455 nm < 325 nm