TE Enriched Sample (E)

The quantum theory of light 29 1.2 A photocell is connected to a variable voltage source of voltage V . (a) The graph below shows how the photocurrent I P changes with V when the photocell is illuminated by 350 nm ultraviolet light of different intensities. A positive V means that the photoemissive metal plate has a lower potential than the anode in the photocell. 0 I p / μA – 1.6 A B C V / V (i) Which curve represents the case with highest intensity? Briefly explain your answer. (ii) Explain why the three curves have the same horizontal intercept of −1.6 V. (b) Consider the case with the lowest intensity. Suppose that the light intensity is 0.01 W m −2 and the area of the photoemissive surface is 4 × 10 −4 m 2 . Given hc = 1243 eV nm. (i) Calculate the max. KE of the photoelectrons emitted and the work function of the metal in unit of eV. (ii) How many photons hit the surface per second? Find the maximum photocurrent if one electron is emitted for every 1000 photons hitting the surface. (c) The experiment is repeated using green light and ultraviolet light of 330 nm, keeping the same saturation photocurrent by adjusting the light intensity. The following graphs are obtained. 0 I p / μA V / V X Y The horizontal intercept of one of the curves is less negative. Which curve represents the case with green light? Briefly explain your answer. high frequency light A variable voltage source As mentioned earlier, for a given opposing voltage V (e.g. −1.2 V), only those photoelectrons having KE higher than the energy barrier (1.2 eV) can reach the anode and form a photocurrent. If V is tuned more negative (i.e. higher energy barrier), the photocurrent becomes smaller. When V goes beyond the stopping voltage (−1.6 V), all photoelectrons are repelled and no photocurrent is produced. Effects on the photocurrent Example 1.6 (Ref: DSE 2016 Q2ai) ( SQ: DSE 2013, 2016, 2020) Teaching notes Guide Ss to revisit the photocell experiment (Fig. 1.7 on p. 7) under the light of Einstein equation . To keep things simple, we shall omit the subscript ‘max’ here. 1. Make a clear distinction between these: K 0 = max. KE of photoelectrons just ejected from cathode K 1 = max. KE of photoelectrons reaching the anode 2. Einstein equation is about K 0 = hf − φ = eV s (not K 1 ). Note that K 0 is independent of the opposing voltage applied V' . 3. If the anode is negative, it repels photoelectrons and slows them down, as they approach the anode. Thus, K 1 = K 0 − eV' = e ( V s − V' ), cutting off at V' = V s . 4. If the anode is positive, it attracts photoelectrons and speeds them up. Thus, K 1 = K 0 + eV' = e ( V s + V' ). 5. Only when no voltage is applied, the anode is neutral and K 1 = K 0 = eV s . (Ref: DSE 2015 Q2.5, 2017 Q2.2) Sample © United Prime Educational Publishing (HK) Limited, Pearson Education Asia Limited 2023 All rights reserved; no part of this publication may be reproduced, photocopied, recorded or otherwise, without the prior written permission of the Publishers.