Physics Paper 2, Nov/Dec. 2010
 Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Main
Weakness/Remedies
Strength

Question 15
Part II : Candidates are expected to answer any three form this part

(a) Explain:
(i) stopping potential;
(ii) work function.

(b)          State two factors that determine the kinetic energy of ejected electrons in photoelectric

(c) State two uses of a photocell.

(d) When the surface of a certain metal is illuminated with yellow light of wavelength 5.893 x 10-7m, electrons of stopping potential 0.36 V are liberated. When the surface is illuminated with violet light of wavelength 3.969 x 10-7 m the stopping potential is 1.38 V.
Calculate the
(i)            Planck's constant, h;
(ii)           Work function, w;
(iii)          Threshold frequency f0,

_____________________________________________________________________________________________________
Observation

(a) Majority of the responding candidates presented the correct explanation of work function but fumbled with that of stopping potential.
(b) The question was found very tractable by most of the respondent candidates.
(c) Two uses of a photocell were correctly stated by majority of the candidates
(d) Few candidates responded satisfactorily.

(a) (i) Stopping potential is the (negative) minimum potential difference of the anode which is required to stop the most energetic electron from reaching the anode. Any valid additional information e.g Unit is volt.
(ii) Work function is the minimum amount of energy required to liberate/eject an electron from the surface of a metal when irradiated by electromagnetic radiation. Any valid additional information e.g. joule.

(b)   Factors affecting kinetic energy of ejected electrons
- work function of surface/metal irradiated

(c)          Uses of photocell
- burglar alarms
- switches for street lights
- automatic doors
- television cameras
- solar calculators
- counters
Accept any other valid example

(d)          Work function = Wo = E - eVs or = hf/λ - eVs

= hc/ - eVs
= hc/5.893 x 10 -7 - 0.36e ..............(α)
= hc/5.969 x 10-7 -1.38e.............(β)

Equation (β) - Equation (α) gives

1.38e - 0.36e = hc/5.969 x 10-7 - hc/5.893 x 10 -7
1.02e = hc (0.2520 - 0.1697) x 107
Taking 1.6 x 10-19 as e and 3.0 x 108 as c,
1.02 x 1.6 x 10-19 = h x 3 X 10-8 x 0.0823 X 107
h = 6.61 X 10-34 Js

From equation (α)
Wo = 6.61 X 10-34x 3.0 X 108
5.893 X 10-7                                    - 0.36 x 1.6 x 10-19
= 3.365 X 10-19      - 0.576 x 10-19
= 2.79 X 10-19 J

= 2.79 X 10-19
6.61 X 10-34
= 4.22 x 10-14 Hz