| Date | May 2011 | Marks available | 2 | Reference code | 11M.2.HL.TZ2.12 |
| Level | Higher level | Paper | Paper 2 | Time zone | Time zone 2 |
| Command term | Describe | Question number | 12 | Adapted from | N/A |
Question
Photoelectric effect and de Broglie wavelength
The diagram is a representation of apparatus used to study the photoelectric effect.
Light from the monochromatic source is incident on a cathode placed in an evacuated tube. A variable voltage supply is connected between anode and cathode and the photoelectric current is registered by the microammeter. The sketch graph shows how the photoelectric current I varies with the potential difference V between anode and cathode for two sources of light, A and B, of different frequencies and intensities.
The frequency of the incident light is increased but the intensity remains constant. Explain why this increase in frequency results in a change to the maximum photoelectric current (saturation current).
The electrons emitted from the photo-cathode have an associated de Broglie wavelength. Describe what is meant by the de Broglie wavelength.
Markscheme
Look for these main points.
light consists of photons whose energy depends on the frequency/hf;
hence the energy available to the (photo)electrons will depend on f;
the potentials VA and VB correspond to/are a measure of the maximum kinetic of the emitted electrons;
the work function (of metal)/energy to emit electron is same for both light sources;
as electrons in A have more kinetic energy available, this frequency must be higher;
(so A)
(i) 1.6 eV ; (answer must be expressed in eV)
(ii) energy of photons = \(\left( {\frac{{6.6 \times {{10}^{ - 34}} \times 8.8 \times {{10}^{14}}}}{{1.6 \times {{10}^{ - 19}}}} = } \right)3.6\left( {{\rm{eV}}} \right)\);
work function=(3.6−1.6=) 2.0eV;
Allow answer in J if (b)(i) expressed in joule (ECF), otherwise award [1 max].
photon energy increases (because frequency increases);
so for same intensity fewer photons per second;
so current reduced / fewer electrons emitted per second;
all particles/electrons exhibit wave properties/have an associated wavelength (called the de Broglie wavelength);
the wavelength is equal to the Planck constant divided by the momentum of the particle/electron/ \(\lambda = \frac{h}{p}\) with terms defined; { (terms must be defined for mark)
Examiners report
i) This was commonly correct but often expressed in joule rather than eV as demanded by the question.
(ii) Again, units were often inappropriate but credit was given if the earlier unit in (b)(i) was incorrect. Many were able to manipulate Einstein’s equation with ease.
Almost all candidates suggested that, in the photoelectric effect, when the frequency of incident light increases but the intensity remains constant, then the maximum emitted current increases. They neglected the dependence of the energy of the photon on its frequency. This is further evidence of the lack of understanding by candidates with this area of the syllabus.
Candidates often described what the de Broglie wavelength is, or gave an equation for it, but rarely both (as the markscheme and the mark allocation required).
