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.

reference to photon
OR
energy = hf or =\(\frac{{hc}}{\lambda }\)

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
\(\frac{{hc}}{{\lambda e}} = \) «\(\frac{{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{40 \times {{10}^{ - 9}} \times 1.6 \times {{10}^{ - 19}}}} = \)» 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.