EJC 2023 J2 H2 PRELIM P3 MS

©EJC 2023 9749/J2H2PRELIM/2023 EUNOIA JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATIONS 2023 9749 PHYSICS MARK SCHEME Paper 3 – Longer Structured Qns Answer Marks 1a Correct positions of max and min intensities, and gradient at 0 and 180 should be approx zero [B1] Max intensity of Io/4 [B1] Working: Io = Io cos2 cos2(90−) = Io cos2 sin2 = Io/4 sin22 [note: sin2 = 2sin cos] 1bi d = 10×10-6  7 = 1.43 × 10−6 m M1 A1 1bii For number of images on each side, sin = 45 nmax =d sin45/ = 1.43 × 10−6 × sin45/ 600×10−9 = 1.68  1 Total number of images = 1 + 1 + 1 = 3 C1 A1 1biii For the first order maxima, 𝑑 sin 𝜃 = 𝜆. Most of the line spacing in the grating is uniform such that the rays converge at  to give a bright image. Some of the line spacing is smaller than the standard spacing , giving rise to some part of the image being formed at a larger angle between  and . B1 B1 180 135 90 45 0

2 ©EJC 2023 9749/J2H2PRELIM/2023 1ci The stationary wave is formed when the incident wave is reflected at the sea wall such that the incident and reflected waves are of the same type, amplitude, frequency, wavelength and speed the incident and reflected wave travelling in opposite directions then overlap and superpose. B1 B1 1cii For the increase in amplitude at wall, the incident and reflected waves must have amplitudes of the same sign / crest meets with crest. OR Constructive interference The reflected wave must have the same phase as the incident wave, hence no phase change occurred during reflection. M1 A1 1ciii 1.75 = 0.78  = 0.78/1.75 v = f = 720 × 0.78/1.75 = 321 C1 A1

3 ©EJC 2023 9749/J2H2PRELIM/2023 Qns Answer Marks 2(ai) The increase in internal energy of a system is the sum of external work done on the system and the heat supplied to the system. B1 B1 2(aii) When roasting, heat is supplied to the potato, thus toQ is positive. onW is negligible since there is little change in volume of the potato) / onW is positive since cooked potato becomes smaller in volume Therefore, increaseΔU is positive. OR onW is negative but magnitude of Q is larger than magnitude of W Therefore, increaseΔU is positive. ΔU=ΔKE+ΔPE ΔPE is negligible as there is no change in state of the potato, therefore, ΔKE is positive. OR ΔPE is positive as water from the potato is changed to gaseous state but ΔKE is also positive. Since KE is directly proportional to (thermodynamic) temperature, the temperature of the potato increases. B1 B1 B1 2(b)(i) ΔU=Q+W Constant volume, therefore W = 0 J ΔU=Q = -55 J B1 A1 2(b)(ii) Process A to B takes place at constant temperature, therefore UA = UB => UAB = 0 For cyclic process, UABCA = 0 UAB + UBC + UCA = 0 UCA = 55 Process C to A is an adiabatic process, therefore toQ = 0 J UCA = Won = 55 Wby = −55 B1 B1 A1

4 ©EJC 2023 9749/J2H2PRELIM/2023 Qns Answer Marks 3(a) r.m.s. value of an alternating current is the value of a steady direct current that will dissipate thermal energy at the same average rate as the a.c. in a given resistor A1 3(b) peak voltage = 6.0 2 peak voltage = 8.5 V A1 3(c) 2 2 (50) 310kf   = = = = Unit: radian per second, rad s−1 (or s-1) A1 A1 3(d) 8.5 V A1 3(e) zero because either no current in circuit (and V = IR) or all p.d. across diode A1 3(f) 2 2 max 8.5 14.5 W5.0 oVP R= = = 11 0.020 s50T f= = = Shape and showing one complete cycle Labelling max power as 14.5 W Labelling period as 0.020 s M1 A1 A1 power/ W t/ s 0 14.5 0.020

5 ©EJC 2023 9749/J2H2PRELIM/2023 Qns Answer Marks 4(a)(i) A beam of white light consists of photons of all wavelengths () in the visible spectrum. The photons interact with electrons in the gas atoms OR The photons will be absorbed by the electrons (Option 1) Photon energy causes electron to move to higher energy level (or to be excited) The photon energy (of certain wavelengths ( E = hf = hc/)) is equal to the energy difference between the energy levels of the gas atoms When electrons de-excite, photons are re-emitted in all directions (Option 2) This results in fewer photons of these wavelengths /photons with much less intensity in the direction of the incident light /original direction of travel. Hence, there will be dark lines corresponding to these wavelengths in the diffraction pattern. B1 B1 B1 B1 B1 4(a)(ii) Transitions −3.40eV to −0.85eV and −3.40eV to −1.51eV B1 4(b)(i) f increases, so energy carried by each photon (= h f) increases. P is constant, so the total incident energy per unit time is constant. Hence, the number of photons per unit time must decrease. The number of photoelectrons must also decrease. no. of photoelectrons emittedquantum yield no. of incident photon= is a fixed value for a given wavelength and type of metal. Correct shape – B1 Indication of fo – B1 4(b)(ii) Negative potential diffrence slows down the electrons, so fewer electrons can reach the collector, photocurrent decreases, to 0 if electrons with max KE cannot reach collector. Positive potential difference accelerates the electrons towards the collector, so all electrons reach the collector, saturation current is reached. B1 I f I V fo

6 ©EJC 2023 9749/J2H2PRELIM/2023 4(b)(iii) With energy per photon (= hf) constant, the number of incident photons increases linearly with power. Hence, the number of photoelectrons also increases linearly with power. B1 4(c)(i) 10 7no. of electrons 6 6.05 10= =  B1 4(c)(ii) 6 13 19 9.2 10no. of electrons per unit time from 10th dynode 5.75 101.6 10 − − = =  13 5 7 5.75 10no. of electrons per unit time from cathode 9.50 106.05 10 = =   M1 A1 4(c)(iii) 56no. of incident photons per unit time 9.5 0 10 3 2.85 10=   =  34 8 6 12 9 power of incident light 6.63 10 3.0 102.85 10 1.57 10 W361 10 hcn  − − − =   =   =   M1 A1 4(c)(iv) 19 18kinetic energy of an electron 1.6 10 50 8.0 1 0 JqV −−= =   =  2 18 31 18 24 8.0 10 J2 2 2 9.11 10 8.0 10 3.82 10 N s pE m p mE − − − − = =   = =     =  34 10 24 6.63 10 1.74 10 m3.82 10 h p − − − = = =   M1 M1 A1 I P

7 ©EJC 2023 9749/J2H2PRELIM/2023 Qns Answer Marks 5ai Binding energy of Mo = 95 × 8.09 = 768.55 MeV Binding energy of La = 139 × 7.92 = 1100.88 MeV Energy released = Binding energy of Mo + Binding energy of La – Binding energy of U Therefore 182 = 768.55 + 1100.88 - Binding energy of U Binding energy of U = 1687.43 MeV = 1690 MeV C1 (either one) M1 A1 5aii Binding energy = Δm c2 Δm = ( ) 6 19 28 1687.43 10 1.6 10 3.0 10 −    273.00 10 −= kg C1 A1 5bi 42 42 0 19 20 1K Ca e −→ + + Also accept: “neutrino”, “anti-neutrino” and 𝜈 A1 5bii From Fig. 5.2, time taken for number of nuclei to drop from N0 to 0.25 N0 is 25 hours. 12 12 2 25 12.5 hours t t = = A1 5biii Graph of Ca is a reflection of the graph for K in the N =0.5 N0 line. Graphs will intersect at N = 0.5 N0 P1 5biv When N (Ca) = 4 N (K), N (K) = 0.2 N0 From the graph, this occurs when t = 29 hours M1 A1 0.0 0.2 0.4 0.6 0.8 1.0 0 5 10 15 20 25 30 35 40

8 ©EJC 2023 9749/J2H2PRELIM/2023 Qns Answer Marks 6(a) acceleration proportional to displacement (from a fixed point) either acceleration and displacement in opposite directions or acceleration always directed towards a fixed point B1 B1 6(b)(i) g and r are constant