Describe the concept of a photon.

In the photoelectric effect there exists a threshold frequency below which no emission of photoelectrons takes place.

Outline how the

(i) wave theory of light is unable to account for this observation.

(ii) concepts of the photon and work function are able to account for this observation.

Light of wavelength 420 nm is incident on a clean metal surface. The work function of the metal is 2.0 eV.

Determine the

(i) threshold frequency for this metal.

(ii) maximum kinetic energy in eV of the emitted electrons.

light consists of discrete packets/quanta/bundles of energy/particle;
each photon has an energy of hf (where h is the Planck constant and f is the frequency of light);

(i) the energy of a (em) wave depends on amplitude (not frequency);
so increasing the intensity should have resulted in electrons being emitted (at any frequency) / OWTTE;

(ii) the work function is the minimum energy required to eject an electron from a metal surface;
if the photon energy (hf ) is less than the work function then no emission will take place;

(i) recognizes that work function = h×threshold frequency;

\({f_0} = \left( {\frac{{2.0 \times 1.6 \times {{10}^{ - 19}}}}{{6.6 \times {{10}^{ - 34}}}} = } \right)4.8 \times {10^{14}}{\rm{Hz}}\);

(ii) recognize that maximum KE=hf-hf₀ or hf-Φ;

\({f_0} = \left( {\frac{c}{\lambda } = \frac{{3.0 \times {{10}^8}}}{{4.2 \times {{10}^{ - 7}}}} = } \right)7.14 \times {10^{14}}{\rm{Hz}}\);
\(hf\left( {{\rm{eV}}} \right) = \left( {\frac{{6.6 \times {{10}^{ - 34}} \times 7.14 \times {{10}^{14}}}}{{1.6 \times {{10}^{ - 19}}}} = } \right)2.96{\rm{eV}}\);

max KE=(2.96–2.0=)0.96eV;