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?

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

              632 nm         <              455 nm      <           325 nm

Explanation:

The equation for the photoelectric effect is

E = hf = Φ + KE  

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

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