Monochromatic light of different frequencies is incident on a metal surface placed in a vacuum. As the frequency is increased a value is reached at which electrons are emitted from the surface. Below this frequency, no matter how intense the light, no electrons are emitted. Outline how the

(i) wave theory of light is unable to account for these observations.

(ii) Einstein model of the photoelectric effect is able to account for these observations.

The graph shows how the maximum kinetic energy E_K of the ejected electrons in (a) varies with the frequency f of the incident light.

Use the graph to determine the

(i) Planck constant.

(ii) work function of the metal.

Show that electrons of energy 0.50 eV have a de Broglie wavelength of about 1.7×10^–9m.

(i) electrons in the metal require a minimum amount of energy to be ejected from the metal;
(according to wave theory) the energy of a wave is dependent on intensity and not frequency;
so given enough time to absorb energy electron emission should take place at any frequency no matter what the intensity / OWTTE;

(ii) photons have energy hf/proportional to frequency (of the light);
an electron may be ejected if this energy is equal to or greater than a threshold value/work function;
the intensity determines the rate of release of photoelectrons, but not their energy;

(i) recognize that slope of graph\( = \frac{h}{e}\) or h (in eV s);
evidence of finding slope eg. \(\frac{{0.5}}{{[6.8 - 5.6] \times {{10}^{14}}}} = 4.17 \times {10^{ - 15}}\); } (accept values in the range of 4.0 to 4.2×10^-15)
\(h = 1.6 \times {10^{ - 19}} \times 4.17 \times {10^{ - 15}} = 6.7 \times {10^{ - 34}}\left( {{\rm{Js}}} \right)\); } (accept values in the range of 6.4 and 6.7×10^-34(Js))
Award [0] for an unsupported correct answer.

(ii) threshold frequency=5.6×10¹⁴(Hz);
work function (hf₀)=6.63×10^-34×5.6×10^-14=3.7×10^-19(J) or 2.3(eV);
If necessary award [2] for use of ECF value of h from (b)(i).
Award [2] for use of any data point and W=hf-Ek giving an answer of 3.7(±0.1)×10^-19(J).
Award [2] for a bald correct answer.

use \(p = \frac{h}{\lambda }\) and \({E_{\rm{K}}} = \frac{{{p^2}}}{{2m}}\) to show that \(\lambda  = \frac{h}{{\sqrt {2m{E_{\rm{K}}}} }}\); (allow equivalent working)
electron kinetic energy=0.5×1.6×10^-19(J) or 8.0×10^-20(J);
\(\lambda  = \left( {\frac{{6.63 \times {{10}^{ - 34}}}}{{\sqrt {2 \times 9.11 \times {{10}^{ - 31}} \times 8.0 \times {{10}^{ - 20}}} }} = } \right)1.74 \times {10^{ - 9}}\left( {\rm{m}} \right)\); } (must see to three significant figures or better)
(≈1.7×10^-9m)
Award marks for evidence of valid working, as the answer is given in the question.