Date | November 2016 | Marks available | 6 | Reference code | 16N.2.HL.TZ0.11 |
Level | Higher level | Paper | Paper 2 | Time zone | 0 - no time zone |
Command term | Calculate, Determine, and Draw | Question number | 11 | Adapted from | N/A |
Question
An apparatus is used to investigate the photoelectric effect. A caesium cathode C is illuminated by a variable light source. A variable power supply is connected between C and the collecting anode A. The photoelectric current I is measured using an ammeter.
A current is observed on the ammeter when violet light illuminates C. With V held constant the current becomes zero when the violet light is replaced by red light of the same intensity. Explain this observation.
The graph shows the variation of photoelectric current I with potential difference V between C and A when violet light of a particular intensity is used.
The intensity of the light source is increased without changing its wavelength.
(i) Draw, on the axes, a graph to show the variation of I with V for the increased intensity.
(ii) The wavelength of the violet light is 400 nm. Determine, in eV, the work function of caesium.
(iii) V is adjusted to +2.50V. Calculate the maximum kinetic energy of the photoelectrons just before they reach A.
Markscheme
reference to photon
OR
energy = hf or =
violet photons have greater energy than red photons
when hf > Φ or photon energy> work function then electrons are ejected
frequency of red light < threshold frequency «so no emission»
OR
energy of red light/photon < work function «so no emission»
i
line with same negative intercept «–1.15V»
otherwise above existing line everywhere and of similar shape with clear plateau
Award this marking point even if intercept is wrong.
ii
«» 3.11 «eV»
Intermediate answer is 4.97×10−19 J.
Accept approach using f rather than c/λ
«3.10 − 1.15 =» 1.96 «eV»
Award [2] for a bald correct answer in eV.
Award [1 max] if correct answer is given in J (3.12×10−19 J).
iii
«KE = qVs =» 1.15 «eV»
OR
1.84 x 10−19 «J»
Allow ECF from MP1 to MP2.
adds 2.50 eV = 3.65 eV
OR
5.84 x 10−19 J
Must see units in this question to identify energy unit used.
Award [2] for a bald correct answer that includes units.
Award [1 max] for correct answer without units.