Question

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?

Answer 1

Answer:

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