A2 Discovery of the Electron
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A2 Discovery of the Electron 155 minutes 143 marks Q1. The diagram shows a narrow beam of electrons directed at right angles into a uniform electric field between two oppositely-charged parallel metal plates at a fixed potential difference. (a) The electrons enter the field at O and leave it at P. Sketch the path of the beam from O to P and beyond P. (2) (b) A uniform magnetic field is applied to the beam perpendicular to the electric field and to the direction of the beam. The magnetic field reduces the deflection of the beam from its initial direction. (i) Explain why the magnetic field has this effect on the beam. ............................................................................................................. ............................................................................................................. ............................................................................................................. (ii) The magnetic flux density is adjusted until the beam passes through the two fields without deflection. Show that the speed v of the electrons when this occurs is given by v= where E is the electric field strength and B is the magnetic flux density. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. (5) (c) In an experiment to measure the specific charge of the electron, electrons were accelerated from rest through a potential difference of 2900 V to a speed of 3.2 × 107 m s–1. Use this information to calculate the specific charge of the electron. ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... (3) (Total 10 marks) Q2. A narrow beam of electrons at a speed of 3.2 × 107 m s–1 travels along a circular path in a uniform magnetic field of flux density, B, as shown in the diagram. (a) Explain why the path of the beam in the field is circular. You may be awarded marks for the quality of written communication in your answer. ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... (3) (b) (i) Show that the speed, v, of the electrons in the field is given by where r is the radius of the circular path of the beam in the field. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. (ii) When the flux density was 7.3 mT, the radius of the circular path of the beam in the field was 25 mm. Use the data to calculate the specific charge of the electron. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. (5) (Total 8 marks) Q3. In an experiment to measure the charge of an oil droplet, a positively charged oil droplet was held stationary by means of a uniform electric field of strength 4.9 × 105 V m–1. (a) (i) What was the direction of the electric field? ............................................................................................................. (ii) Show that the specific charge of the oil droplet was 2.0 × 10–5 C kg–1. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. (3) (b) When the electric field was switched off the oil droplet fell and quickly reached constant speed.Explain why the oil droplet reached constant speed. ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... (3) (Total 6 marks) Q4. Electrons are emitted by the process of thermionic emission from a metal wire in anevacuated container. The electrons are attracted to a metal anode which has a small hole at its centre. The anode is at a fixed positive potential relative to the wire. A beam of electrons emerges through the hole at constant velocity. (a) Explain (i) what is meant by thermionic emission, ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. (ii) why it is essential that the container is evacuated, ............................................................................................................. ............................................................................................................. (iii) why the anode must be at a positive potential. ............................................................................................................. ............................................................................................................. (4) (b) An electron is accelerated from rest through a potential difference of 2500 V between the wire and the anode. Calculate (i) the kinetic energy of the electron at the anode, ............................................................................................................. ............................................................................................................. (ii) the speed of the electron at the anode. Ignore relativistic effects. ............................................................................................................. ............................................................................................................. ............................................................................................................. (4) (Total 8 marks) Q5. In an experiment to measure the charge on a charged oil droplet, a droplet was observed between two horizontal metal plates, as shown in the diagram below, spaced 6.0 mm apart. (a) The oil droplet was held stationary when a negative potential of 320 V was applied to the top plate, keeping the lower plate at zero potential. (i) State the sign of the charge on the droplet. ............................................................................................................. (ii) With reference to the forces acting on the droplet explain why it was stationary. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. (2) (b) The potential difference between the plates was then switched off and the droplet fell at constant speed through a vertical distance of 1.20 mm in 13.8 s. (i) Calculate its speed of descent. ............................................................................................................. ............................................................................................................. (ii) By considering the forces on the spherical droplet of radius r as it falls at constant speed v, show that where is the viscosity of the air between the plates and ρ is the density of the oil. Ignore buoyancy effects. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. (iii) Calculate the radius of the droplet and hence show that its mass is 2.6 × 10–15 kg. viscosity of the air = 1.8 × 10–5 N s m–2 density of the oil = 960 kg m–3 ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. (iv) Calculate the charge carried by this droplet. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. (10) (Total 12 marks) Q6. A potential difference was applied between two electrodes in a glass tube containing air, as shown in the diagram below. The pressure of the air in the tube was gradually reduced until a glow of light was observed between the electrodes. (i) Explain why light was emitted. ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... (ii) State why the glow was not observed until the pressure of the air in the tube was low enough. ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... (Total 4 marks) Q7. A charged oil droplet was observed falling between two oppositely charged parallel plates, as shown in Figure 1. Figure 1 (a) Explain why the droplet stopped moving and remained stationary when the potential difference between the plates was adjusted to a certain value, Vc. ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... (3) (b) (i) The spacing between the plates is 6.0 mm. A charged oil droplet of mass 6.2 × 10–14 kg is stopped when Vc = 5700 V. Calculate the charge on this droplet. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. (ii) Describe and explain what would have happened to this droplet if the potential difference had been greater than 5700 V. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. (5) (Total 8 marks) Q8. A narrow beam of electrons is produced in a vacuum tube using the arrangement shown inFigure 1. Figure 1 (a) Describe the function of each voltage supply unit and state a typical voltage for each unit. (i) unit A ............................................................................................................. ............................................................................................................. (ii) unit B ............................................................................................................. ............................................................................................................. (3) (b) State and explain the effect on the beam of (i) reducing the voltage of A, ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. (ii) increasing the voltage of B. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. (4) (Total 7 marks) Q9. A narrow beam of electrons is directed into a uniform electric field created by two oppositely charged parallel horizontal plates, as shown in the figure below. The initial direction of the beam is perpendicular to the direction of the electric field. The beam makes a visible trace on a vertical fluorescent screen. (a) Explain why the beam curves upwards at an increasing angle to the horizontal. You may be awarded marks for the quality of written communication in your answer. ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... (4) (b) When a uniform magnetic field of a certain flux density is applied perpendicular to the screen, the beam passes between the plates undeflected. (i) Show that the beam is undeflected when the magnetic flux density B = , where E is the electric field strength between the plates and v is the speed of the electrons. ............................................................................................................. ............................................................................................................. ............................................................................................................. (ii) Hence show that the specific charge, e/m, of the electron can be calculated using where VA is the anode voltage and B is the magnetic flux density needed for zero deflection. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. (iii) Determine the specific charge of the electron using the following data: anode voltage the plates = 3800 Vplate separation mmmagnetic flux density = 4500 Vpotential difference between = 50 = 1.9 mT ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. (7) (Total 11 marks) Q10. The figure below shows an electron gun in an evacuated tube. Electrons emitted bythermionic emission from the metal filament are attracted to the metal anode which is at a fixed potential, V, relative to the filament. Some of the electrons pass though a small hole in the anode to form a beam which is directed into a uniform magnetic field. (a) (i) Explain what is meant by thermionic emission. ............................................................................................................. ............................................................................................................. (ii) Show that the speed,v, of the electrons in the beam is given by where m is the mass of the electron and e is the charge of the electron. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. (3) (b) The beam of electrons travels through the field in a circular path at constant speed. (i) Explain why the electrons travel at constant speed in the magnetic field. ............................................................................................................. ............................................................................................................. ............................................................................................................. (ii) Show that the radius, r, of the circular path of the beam in the field is given by where B is the magnetic flux density and V is the pd between the anode and the filament. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. (iii) The arrangement described above was used to measure the specific charge of the electron, e/m. Use the following data to calculate e/m. B = 3.1 mTr = 25 mmV = 530 V ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. (7) (Total 10 marks) Q11. (a) The diagram below shows a narrow beam of electrons produced by attracting electrons emitted from a filament wire to a metal plate which has a small hole in it. (i) Why does electric current through the filament wire cause the wire to emit electrons? ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. (ii) Why must the filament wire and the metal plate be in an evacuated tube? ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. (3) (b) The voltage between the filament wire and the metal plate is 3900 V. For each electron emerging through the hole in the plate, calculate (i) the kinetic energy, in J, ............................................................................................................. ............................................................................................................. (ii) the speed. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. (4) (Total 7 marks) Q12. A narrow beam of electrons is produced in a vacuum tube using an electron gun, part of which is shown in Figure 1. Figure 1 (a) (i) State and explain the effect on the beam of electrons of increasing the filament current. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. (2) (ii) State and explain the effect on the beam of electrons of increasing the anode potential. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. (2) (b) The beam of electrons is directed at right angles into a uniform magnetic field as shown inFigure 2. Figure 2 (i) Explain why the electrons move in a circular path at a constant speed in the magnetic field. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. (3) (ii) When the speed of the electrons in the beam is 7.4 × 106 m s–1 and the magnetic flux density is 0.60 m T, the radius of curvature of the beam is 68 mm. Use these data to calculate the specific charge of the electron, stating an appropriate unit. Give your answer to an appropriate number of significant figures. answer = ...................................... (4) (iii) Discuss the historical relevance of the value of the specific charge of the electron compared with the specific charge of the H+ ion. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. (2) (Total 13 marks) Q13. In an experiment to measure the charge of the electron, a charged oil droplet of unknown mass was observed between two horizontal parallel metal plates, as shown in the figure below. (a) The droplet was observed falling vertically at its terminal speed when the pd between the plates was zero. (i) By considering the forces acting on the droplet as it falls at its terminal velocity, v,show that the radius, r, of the droplet is given by where η is the viscosity of air and ρ is the density of the oil droplet. (2) (ii) Explain how the mass of the oil droplet can be determined from its radius, r. ............................................................................................................. ............................................................................................................. ............................................................................................................. (1) (b) (i) The two horizontal parallel metal plates were 5.0 mm apart. The mass of the droplet was 6.8 × 10–15 kg. The droplet was held stationary when the plate pd was 690 V. Calculate the charge of the oil droplet, expressing your answer to an appropriatenumber of significant figures. answer.......................................... C (3) (ii) Millikan made the first accurate measurements of the charge carried by charged oildroplets. Outline what Millikan concluded from these measurements. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. (2) (Total 8 marks) Q14. In the figure below, a beam of monoenergetic electrons is produced by thermionic emission from a wire filament in an evacuated tube. The beam is directed at a thin metal sample at normal incidence and it emerges from the sample in certain directions only, including its initial direction. (a) (i) Name the physical process occurring at the thin metal sample in the figure above which shows the electrons behaving as waves. ............................................................................................................. ............................................................................................................. (1) (ii) Explain why the electrons need to be monoenergetic in order for them to emerge incertain directions only. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. (2) (b) A transmission electron microscope (TEM) operating at an anode potential of 25kV isused to observe an image of a thin sample. (i) Calculate the momentum of the electrons emerging from the anode, stating anappropriate unit. answer = ......................................................... (4) (ii) Describe and explain how the resolution of the image would change if the anodepotential were increased. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. (3) (Total 10 marks) Q15. A narrow beam of electrons is directed into the region between two parallel plates, P andQ. When a constant potential difference is applied between the two plates, the beam curves downwards towards plate Q as shown in the figure below. (a) Explain why the beam curves downwards at an increasing angle to its initial direction. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ (3) (b) A uniform magnetic field is then applied at right angles to both the beam and the electric field between the plates P and Q. As a result, the downward deflection of the beam is increased. (i) The arrangement is to be used to determine the speed of the electrons in the beam. Describe what adjustments to the flux density B of the magnetic field should be made to reduce the deflection of the beam to zero. ............................................................................................................... ............................................................................................................... ............................................................................................................... (1) (ii) Explain why the electrons pass undeflected through the fields when their speed v is given by where V is the potential difference between plates P and Q and d is the perpendicular distance between the plates. ............................................................................................................... ............................................................................................................... ............................................................................................................... ............................................................................................................... ............................................................................................................... ............................................................................................................... (2) (c) The beam of electrons was produced by thermionic emission from a heated filament. When the potential difference between the anode and the filament was 4200 V, the speed of the electrons in the beam was 3.9 × 107 ms–1. Use this information to determine the specific charge of the electron. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ answer = ............................. C kg–1 (3) (Total 9 marks) Q16.A charged oil droplet was observed between two horizontal metal plates X and Y, as shown in the diagram below. (a) (i) With the switch S open, the droplet fell vertically at a constant velocity of 1.1 × 10−4ms−1. Show that the radius of the droplet is about 1.0 × 10–6 m. Assume the droplet is spherical. density of oil, ρ = 880 kg m−3viscosity of air, η = 1.8 × 10−5 N s m−2 (4) (ii) Calculate the mass of the droplet. mass .......................................... kg (1) (iii) The switch S was closed and the potential difference from the voltage supply was adjusted gradually to reduce the downward motion of the droplet. The droplet stopped moving when the potential difference across the plates was 680 V. The spacing between the plates was 6.0 mm. Calculate the magnitude of the charge on the droplet. charge ........................................... C (3) (b) The mass of another charged droplet was found to be 4.3 × 10−15 kg. With switch S closed and the voltage supply at its maximum value of 1000 V, this droplet fell more slowly than when the switch was open but it could not be stopped. Explain why this droplet could not be held at rest and show that the magnitude of the charge on it was 1.6 × 10−19 C. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ (4) (Total 12 marks) M1. (a) path curves upwards from O to P path is tangential to curve at P and straight beyond P 2 (b) (i) magnetic field exerts a force on a moving charge/electron (1) magnetic force has a downwards component (at all points) [or magnetic force < electric force] (1) (ii) magnetic force = Bev (1) electric force Bev = eE (gives v = = eE (1) ) (1) 5 (c) work done (or eV) = gain of kinetic energy (or ½mv2) (1) (1) = 1.8 × 1011 C kg–1 (1) 3 [10] M2. (a) magnetic force perpendicular to (direction of) motion (or velocity) (1) force does not change speed (or force does no work) (1) force causes direction of motion to change (1) force (or acceleration) is centripetal/ acts towards centre of curvature (1) velocity is tangential (1) max 3 QWC 2 (b) (i) magnetic force = Bev (1) centripetal acceleration = , Bev = (1) (gives v = (ii) ) = Bev gives (1) = (1) = 1.75 × 1011 C kg –1 (1) 5 [8] M3. (a) (ii) (i) (vertically) upwards (1) mg = qE, (1) (1) (= 2.0 × 10–5 C kg–1) 3 (b) initial downwards acceleration due to weight (or gravity) (1) viscous force/drag/friction (or resistance) due to air increases with increase in speed (1) speed increases until drag become equal to (and opposite to) weight (no resultant force) hence no acceleration (1) max 3 [6] M4. (a) (i) metal wire emits electrons when heated (1) conduction electrons in metal gain kinetic energy when wire is heated (1) (ii) electrons from wire would be absorbed/scattered/stopped by gas atoms or collide with gas atoms and lose kinetic energy or speed (1) (iii) electrons carry negative charge so anode needs to be positive (to attract them) (1) 4 (b) (i) Ek (or ½mv2) (= work done or eV) = 1.6 × 10–19 × 2500 (1) = 4.0 × 10–16 J (1) (ii) = (1) = 3.0 × 107 m s–1 (1) (allow C.E. for value of Ek from (i)) 4 [8] M5. (a) (ii) (i) positive (1) electric force directed upwards = weight (1)[or = mg] 2 (b) (i) v= = 8.7 × 10–5 m s–1 (1) (ii) weight [or mg] = 4/3 πr3 ρg (1) (since speed constant) viscous force = 6 πηrv (1) 4 /3r3 g = 6πηrv to give desired equation (1) (iii) rearrange equation to give r = (1) = 8.7 × 10–7 m (1) (8.65 × 10–7 m) (allow C.E. for value of v from (i), but not 3rd mark) m (= 4/3 πr3 ρ) = 4/3 π (8.65 × 10–7)3 × 960 (1) (= 2.6 × 10–15 kg) (iv) = mg (1) Q= = 4.8 × 10–19 C (1) (1) (4.78 × 10–19 C) 10 [12] M6. (i) electrons [or ions] present (1) electrons/ions accelerated by electric field [or electrons and ions collide] (1) excitation/ionisation of gas atoms/ions/molecules/particles occur (1) photons emitted on return to lower energy or ground state (1) (ii) electrons/ions do not gain enough kinetic energy (to produce ionisation) (1) because too many atoms/ions/molecules/particles present (1) max 4 QWC 1 [4] M7. (a) force due to electric field is vertically upwards and proportional(or related to) plate pd (1)at V = Vc, force due to field is equal and opposite to the weight ofthe droplet (1)no resultant force (or forces balance) at Vc (droplet remains stationary) (1) 3 (b) (i) electric force (or qV/d) = weight (or mg) (1) q = (1) = 6.4 × 10–19C (1) (ii) for pd > 5700 (V), droplet moves upwards (1) due to increased electric force (1) droplet reaches terminal velocity (1) max 5 [8] M8. (a) (ii) (i) unit A: supplies current/power/energy to the filament or heats the filament (1) 0 – 50 V (1) unit B: to make the anode positive w.r.t. the filament, so that electrons are attracted/accelerated to the anode (1) > 250 V (1) max 3 (b) (i) beam current or intensity is reduced (1) (because) fewer electrons are emitted (per sec) from the filament (1) [or no beam as no electrons emitted if voltage of A reduced enough (1) (only)] (ii) electrons travel faster [or more kinetic energy] (1) (because the force of) attraction to the anode is greater (1) 4 [7] M9. (a) each electron experiences an electrostatic force (vertically) upwards (1)this force does not change as the electron moves across the field (1)each electron (therefore) has a (constant) acceleration vertically upwards (1)velocity of each electron has a constant horizontal component of velocity (1)[or has an increasing vertical component of velocity]so the direction of motion/velocity becomes closer and closer to avertical line (as electron moves across the field) (1)[or angle to the vertical becomes less] Max 4 QWC 1 (b) (i) (for beam to be undeflected) force due to electric field, eE (or qE) (1) equals force due to magnetic field, Bev (1) (gives v = (ii) (k.e. at anode ) = ½mv2 = eVA (1) gives (iii) ) E(= (1) (i.e. = )= ) (1) (= 7.6 × 104 (V m–1)) (1) = 1.8 × 1011 C kg–1 (1) 7 [11] M10. (b) (a) (i) emission of (conduction) electrons from a heated metal (surface) or filament/cathode (1) work done on electron = eV (1) (ii) gain of kinetic energy (or ½ mv2) = eV; rearrange to give required equation (1) (i) 3 work done = force × distance moved in direction of force (1)force (due to magnetic field) is at right angles to the direction ofmotion/velocity[or no movement in the direction of the magnetic force no work done] (1)electrons do not collide with atoms (1) any two (1)(1) [alternative for 1st and 2nd marks(magnetic) force has no component along direction of motion (1)no acceleration along direction of motion (1)or acceleration perpendicular to velocity] (1) (1) (1) (iii) (rearranging the equation gives) (1) Ckg–1 (1) 7 [10] M11. (a) (i) current heats the wire (1) electrons (in filament) gain sufficient k.e. (to leave the filament) (1) (ii) electrons would collide (or be absorbed or scattered) by gas atoms (or molecules) (1) 3 (b) (i) k.e. (= eV) = 1.6 × 10–19 × 3900 (1) = 6.2 × 10–16 (J) (1) (ii) (rearrange ½ m v2 = eV to give) v (= (2eV/m) 1/2) = (1) = 3.7 × 107 m s–1 4 [7] M12. (a) (i) The number of electrons (per second) in the beam will increase (1) because the filament will become hotter and will emit more electrons (per 2 second) (1) 2 (ii) the speed (or kinetic energy) of the electrons will increase (1) because the electrons (from the filament) are attracted towards the anode with a greater acceleration (or force) (1) (or gain more kinetic energy in crossing a greater pd) 2 (b) (i) (magnetic) force on each electron in the beam is perpendicular to velocity (1) no work is done on each electron by (magnetic) force so ke (or speed) is constant (1) magnitude of (magnetic) force is constant because speed is constant (1) (magnetic) force is always perpendicular to velocity so is centripetal (1) max 3 (ii) rearranging r = gives (1) = 1.81 × 1011 (1) C kg–1 (1) for correct answer to 2 sf (1) 4 (iii) specific charge for the electron ≈ 2000 × specific charge of H+ (1) (accept = and accept any value between 1800 and 2000) which was the largest known specific charge before the specificcharge of the electron was determined/measured (1) (or which could be due to a much greater charge or a much smallermass of the electron) 2 [13] M13. (a) (i) either (at terminal speed (v)) the viscous force on the droplet = its weight (or mg or the force of gravity on it) or viscous force = 6π rv (where r is the radius of the droplet and is its viscosity) and weight (= mg) = 4πr3ρg/3 4πr3ρg/3 = 6π rv (which gives r = (9 v/2ρg)½) 2 (ii) r (can be calculated as above then) used in the formula m = 4 πr3ρ/3 to find the droplet mess, m [owtte] alternatively; (from 6π rv = mg) (as all values are known use) m = 6π rv/g 1 (b) (i) electric force (or QV/d) = the droplet weight (or mg) Q= C 2 sf answer 3 (ii) any two from the charge on each droplet is a whole number × 1.6 × 10–19 C(or × charge of the electron) the least amount of charge (or the quantum of charge) is thecharge of the electron the quantum of charge is 1.6 × 10–19 C [owtte] max 2 [8] M14. (a) (i) diffraction 1 (ii) the electrons in the beam must have the same wavelength otherwise electrons of different wavelengths (or speeds/velocities/energies/momenta) would diffract by different amounts (for the same order) [owtte] 2 (b) (i) (eV = ½ m v2 gives) either v = or 1.6 × 10–19 × 25000 = ½ × 9.1 × 10–31 × v2 v= = 9.4 × 107 m s–1 p or mv (= 9.1 × 10–31 × 9.4 × 107) = 8.5 × 10–23 kg m s–1 (or N s) alternatives for first two marks p or mv = = 4 (ii) any two of the first three mark points increase of pd increases the speed (or velocity/energy/momentum) of the electrons (so) the electron wavelength would be smaller (and) the electrons would diffract less (when they passthrough the lenses) and the image would show greater resolution (or be moredetailed) max 3 [10] M15. (a) force due to electric field acts (vertically) downwards on electrons vertical (component) of velocity of each electron increases horizontal (component of) velocity unchanged (so angle to initial direction increases) 3 (b) (i) magnetic flux density should be reversed and adjusted in strength (gradually until the beam is undeflected) 1 (ii) magnetic (field) force = Bev and electric (field) force = eV / d (Accept Q or q as symbol for e (charge of electron) Bev = eV/d (for no deflection) gives v = V / Bd 2 (c) (gain of) kinetic energy of electron = work done by anode pd or ½ m v2 = e V(A) = 1.8 × 1011 C kg–1. 3 [9] M16.(a) (i) (at terminal velocity v ), weight of droplet ( or mg) = viscous drag (or 6 Backward working 3 marks max; viscous force ( = 6 r v ) = 6 × 1.8 × 10−5 × 1.0× 10−6 × 1.1 × 10−4 = 3.7 × 10−14 N mass (m) of droplet = (4 weight = mg = (allow 3.7) r3 / 3 ) × ρ , (where r is the droplet radius ) r v) (therefore) (4 r3 / 3 ) × ρ g = 6 r v ( or rearranged) (hence) r (= (9 ɳ v / 2 ρ g)1/2 (therefore) viscous force = weight as required for constant velocity note; some evidence of calculation needed to give final mark Allow final answer for r in the range 1 to 1.05 × 10−6 to any number of sig figs 4 (ii) Allow ecf for r from a(i) in a correct calculation that gives m in the range 3.6 to 4.0 × 10−15 kg ( or correct calculation of 6 r v / g) 1 (iii) electric force ( or QV / d) = droplet weight ( or mg) Allow ecf m (or r) from a(ii) (or a(i)). Accept values in 1st mark line [or Q (= viscous force × d / V Use of e instead of Q or q = 2 marks max =6 × 1.8 × 10−5× 1.0 × 10−6× 1.1 × 10−4 × 6.0 × 10−4 / 680 ] For the 2nd mark, allow use of viscous force calculation. Use of viscous force method does not get 1st mark. Q = 3.2 × 10−19C If both methods are given and only one method gives Q = ne (where n = integer >1), ignore other method for 2nd mark and 3rdmark. For the final mark, Q must be within n e ± 0.2 × 10 −19 from a correct calculation. 3 (b) The weight of the second droplet is greater than the maximum electric force on it Alternative for 1st mark; weight = drag force + elec force ( owtte) Scheme using V for next 5 marks; If n =1 for the second droplet , pd to hold it = 1580 V ( = mgd / e) which is not possible as V max = 1000 V If n = 2 , it would be held at rest by a pd of 790 V ( = 1580 / 2 or 680 × 4.3 / 3.7 V) if n > 2 , it would be held at rest by a pd of less than 790 V ( or 790 / n V) So n =1(e ) must be the droplet charge Alternative schemes for last 5 marks Q scheme Using QV /d = mg for a stationary droplet gives Q = mgd / V = 2.53 − 10−19 C which is not possible as Q = integer x e (so) Q (=ne) < 2.53 × 10−19 C owtte) Calculation to show Q= 1e fits above condition Q= 2e does not fit above condition F scheme;- Calc of mg to give 4.2 (+0.2) × 10−14 N Calc for Q = 1e of QV / d to give 2.6(+0.2) × 10−14 N Calc for Q =2e of QV / d to give 5.3 (+0.2)× 10−14N mg> elec force for Q =1e or <2e for Q=2e So n =1(e ) must be the droplet charge Max 4 [12] E1. Both marks were usually gained in part (a) and only a minority of candidates continued the curve beyond the metal plates. In part (b)(i) many candidates failed to mention that there is a force on an electron moving through the magnetic field or that, in this case, the force had a downward component. Many candidates in part (ii) were able to state and use the correct equation but only a few candidates stated which force each term stood for. In part (iii), many candidates knew the correct equation but used it to calculate the charge of the electron instead of its specific charge. E2. In part (a) the majority of candidates knew that in the situation described, the force was perpendicular to the direction of motion and was therefore centripetal. Not many candidates however, mentioned that the force causes the direction of motion to change or that it does not change the speed. Only a small minority knew that the velocity is tangential to the path. Many candidates failed to provide an adequate derivation, in part (b)(i), for the speed, even though they produced the required equation. Many candidates scored both marks in part (b)(ii) although a significant minority calculated the charge of the electron rather than its specific charge. E3. In part (a)(i) most candidates were able to give the correct direction of the electric field. In part (ii), a significant minority attempted to calculate the charge of the oil droplet by either considering the mass of the droplet as 1 kg or neglecting the mass altogether. Many other candidates knew that the question required a calculation of the charge/mass ratio for the droplet, but gave their explanation in terms of the charge, e, of the electron instead of the charge on the droplet. When explaining the effect of switching off the electric field in part (b), most candidates knew that the drag force equalled the weight of the droplet at the terminal speed. A significant number of candidates however, did not make it clear that the drag force increased as the speed increased. Few candidates mentioned that the droplet accelerated initially due to its weight or that the acceleration became zero at terminal speed. E4. In part (a)(i), most candidates were aware that the wire needed to be heated in order to free electrons from it but few referred to the kinetic energy gained by conduction electrons. In part (ii), many candidates referred to ionisation of air molecules rather than electrons being stopped or slowed down by air molecules. In part (iii), most candidates knew why the anode needed to be at a positive potential. Most candidates know how to calculate the gain of kinetic energy in part (b)(i) but many candidates lost a mark as a result of a significant figure error. The calculation in part (ii) was usually completed correctly. E5. In part (a) (i), most candidates stated the type of charge correctly and also knew that the weight of the droplet and the electric field force were equal and opposite when the droplet was stationary. Some candidates lost a mark as a result of imprecise references to electric field acting on the droplet. The calculation of the speed of descent of the droplet in part (b) (i), was usually done correctly. In part (ii), most candidates gave a correct expression for the weight of the droplet in terms of its radius and were then able to equate this to the correct expression for the viscous force. In part (iii), most candidates were able to rearrange the equation given in part (ii) correctly and then calculate the radius and mass of the droplet correctly. Some candidates failed to state the correct unit for the droplet radius. Finally, in part (iv), the general principle of the calculation was known by most candidates although some candidates used an incorrect value for the plate spacing. The symbol for electronic charge was frequently used for the droplet charge. E6. In part (i), most candidates were aware that electrons travelled along the tube and many realised that collisions between the gas atoms and the electrons caused excitation or ionisation of the gas atoms. They were also aware that photons were emitted from gas atoms when deexcitation of the atoms occurred. However, candidates often failed to mention that the electrons were accelerated along the tube by the electric field. Although most candidates knew in part (ii) that unless the pressure was sufficiently low that there were too many gas atoms present, very few could then give a clear explanation of why excitation takes place when the number of gas atoms is reduced. E7. In part (a), most candidates were aware that the force of gravity acted downwards and the electric force acted upwards on the droplet, although some candidates considered separate electric forces from each of the parallel plates. However, some candidates failed to indicate that the electric force increased as the plate voltage increased, thereby causing the droplet to become stationary at a particular voltage. Although most candidates realised that the forces were equal at this particular plate voltage, not all of them made clear that the forces were therefore balanced or the resultant force was therefore zero. In the calculation in part (b) (i), most candidates equated the correct expressions for the weight and the electric force and were thus able to calculate the charge on the droplet from data given. Some candidates lost a mark as a result of giving the droplet charge in the equation as e rather than Q or q. A significant minority also lost a mark through omitting the unit in the final answer. In part (ii), most candidates knew that the electric force increased as the plate pd was increased and that consequently the droplet would move upwards. However, very few candidates realised that the droplet would reach terminal velocity in its upward motion. E8. Although many candidates, in part (a) (i) knew the function of the voltage supply (unit A) in the vacuum tube, a significant number failed to mention, in part (ii), that the purpose of the second supply (unit B) was to accelerate or attract electrons to the anode. Many candidates scored a mark in (i) for giving a voltage less than 50 V for supply A, although not as many candidates were able to give a suitable voltage for supply B. In part (b) (i), most candidates knew that reducing the voltage of supply A caused fewer electrons to be emitted from the filament and many were able to deduce that the effect was to produce a less intense beam. In part (ii), although candidates often realised that increasing the voltage of supply B caused a greater force of attraction on the electrons towards the anode, many thought the result was to make the beam more intense rather than to accelerate the electrons to a greater speed. E9. Once again, few candidates were able to develop a coherent explanation in part (a), but most candidates did gain some marks. The majority of candidates were aware that an electron between the plates experienced an upward force due to the electric field in existence between the plates, but weak candidates often referred to the force on an electron due to the positive anode. Few candidates stated that the force on each electron between the plates was constant, and many thought that the upward curve of the beam was due to an increasing force as the electron moved towards the positive plate. Many candidates did state that the upward force caused an upward acceleration. Some candidates correctly realised that the horizontal component of velocity was constant. However, few candidates drew their statements together to relate the vertical and horizontal components of velocity to the upward curvature of the beam. Although many candidates in part (b) (i) correctly equated the expressions for the magnetic and electric forces, many failed to state clearly which expression represented which force. In part (ii), most candidates gave the anode voltage equation correctly and were able to use it to derive the required expression for the specific charge. Candidates who failed to score marks in part (ii) usually introduced irrelevant equations and used them extensively without any success. In part (iii), most candidates correctly calculated the electric field strength and then used it to calculate the specific charge of the electron. Some candidates made careless arithmetical errors in their calculation of the electric field strength and some failed to give the correct unit for the specific charge. Others obtained the correct value for e/m but then lost the final mark as a result of extending the calculation to give e as their answer for the specific charge. E10. In part (a) (i), candidates often failed to score the mark because they did not realise metal needed to be heated in order to emit (thermionic) electrons. In part (ii), most candidates knew the basic derivation of the equation for the speed of the electrons but few scored full marks because they did not state clearly that eF was the work done on each electron by the pd between the filament and the anode. Few satisfactory answers were obtained in part (b) (i). In general, although many candidates were aware that the magnetic field force was perpendicular to the velocity of the electrons, very few realised that the speed of each electron was constant because its acceleration had no component in its direction of motion. In part (ii), most candidates successfully combined the equation r = BQ/mv with the equation given in part (a) (ii) to obtain the required equation. However, some candidates confused v and V and others were unable to carry out the basic algebraic operations needed to combine the two starting equations. In part (iii), most candidates successfully rearranged the equation in part (b) (ii) and then used the data to obtain the correct value for e/m, although some lost the final mark as a result of a unit error. E12. In (a) (i) most candidates understood the heating effect of an electric current, realising the beam would become more intense due to the release of more electrons, but struggled for a precise and rigorous explanation. Many thought the electrons were emitted with increased speed and a significant number of candidates did not explain the effect on the beam, as asked in the question, and simply said more electrons were emitted. In (a) (ii) most candidates identified that the electrons would be travelling at greater speed but again struggled for a complete explanation. In (b) (i) most candidates scored one mark for identifying that the force on the electrons acted at right angles to their direction of motion. More able candidates identified the fact that the force does no work on the electrons and hence their speed remains constant. Very few candidates could go on to say that the force is always at right angles to their direction of motion and hence is centripetal. The calculation in (b) (ii) should have been a straightforward exercise but less than a quarter of the candidates scored all four marks. The necessary equation is on the data sheet or candidates could equate the centripetal force to the force on an electron in a magnetic field. However many candidates did not know what was meant by specific charge and simply worked out the charge on an electron. A significant number of candidates used values of e or m or both from the data sheet in arriving at their value of e/m. This approach received no marks. The significant figure and unit marks were independent. Although the unit of specific charge was widely known (on the Data and Formulae Booklet) a large number of candidates gave their answer to three significant figures even though all the data was presented to two significant figures. In (b) (iii) the candidates that had worked out e in (ii) went on to talk about quantisation of charge and Millikan’s experiment for which they gained no credit. This question is a crucial part of the turning points option and as such was poorly answered. Very few candidates could clearly say that the specific charge of an electron was much greater than that of the H ion which could have been due to either a much greater charge or a much smaller mass of the electron. E13. In part (a)(i), many candidates knew the correct expression for the weight in terms of the droplet radius and equated it to the Stokes’ law expression for the upward force, but they often did not identify the nature of the upward force as a viscous force. Most candidates knew in (a)(ii) that the mass could be calculated from the product of the oil density and the droplet volume and were able to give the correct formula for the droplet volume in terms of its radius. Those candidates who did not know the volume formula were unable to gain this mark. Some candidates obtained the mark by showing the droplet mass could be calculated by dividing the viscous force expression by g. Some candidates lost marks in part (b) because they used the symbol e instead of Q or q for the droplet charge or they gave their answer to three or more significant figures. In part (b)(ii), candidates were often unclear about what the quantum of charge is or they stated that the charge of a droplet is a factor rather than a multiple of the charge of the electron. E14. In part (a) most candidates named the process correctly in part (i) and demonstrated some knowledge of the process in the subsequent parts of the question. In (ii), many candidates knew that electrons with different speeds would not diffract in the same directions although some candidates did not mention that this is because they would have different wavelengths. Many correct calculations were seen in part (b)(i) and most candidates scored the unit mark. Some candidates lost marks through careless numerical errors. In (b)(ii), most candidates knew that the speed would increase and therefore the wavelength would be decrease. However, some candidates thought that this would cause more diffraction or that the resolution would be poorer. Candidates gained no marks if they gave an incorrect physical reason such as a greater number of electrons in the beam. E15. In part (a) most students knew that the beam was attracted towards the positive plate because electrons have a negative charge. Many students realised the electrons were accelerated downwards but failed to recognise either the acceleration was vertically downwards or the vertical component of velocity increased. Relatively few students stated that the horizontal component of velocity was unchanged. A common misconception was that the electrons accelerated downwards because the force on each electron increased as it moved closer to the positive plate. In (b)(i) most students stated either that the magnetic field needed to be adjusted or it needed to be reversed, few students stated the field needed to be reversed and adjusted. In (b)(ii), most students equated the relevant force expressions but a significant number did not identify which expression was which. In part (c) many students scored full marks with a clear and accurate calculation. A significant number of students knew the correct equation to be used but were then unable to gain further credit because they then inserted the value of e or m from the data sheet into the equation to calculate e/m. Students who used a value of e or m from the data sheet to calculate e/m were unable to score beyond the first mark. E16.(a) (b) Many candidates made good progress in this part question and provided clear explanations and calculations. Some candidates lost a mark in ai by not making clear the forces to be equated. In aiii, candidates were generally able to give the correct force equation and then to use it correctly with the given data to calculate the drop charge. Most candidates knew that the droplet’s weight was greater than the electric force. Candidates who realised that this inequality gives an upper limit on the droplet charge generally then made good progress by calculating this upper limit and then using it with their knowledge that the droplet charge is a whole number x e to show that the droplet in question must have a charge of e. Candidates who made progress by calculating the forces involved on droplets carrying charges of 1e or 2e or the potential differences necessary to stop such droplets often failed to make clear what the charge values were in each calculation.
