Oxford Revise AQA A Level Physics | Chapter P27 answers
P27: The discovery of the electron
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Question |
Answers |
Extra information |
Mark |
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Spec reference |
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01.1 |
Anode is positive; cathode is negative |
Both needed for mark |
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3.12.1.1 |
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01.2 |
When a magnet was brought near the tube, the positive column was distorted/deflected |
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3.12.1.1 |
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01.3 |
The applied potential difference is sufficient to produce an electric field that pulls electrons out of atoms and ionises them When the positive ions and electrons recombine, photons are emitted |
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3.12.1.1 |
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01.4 |
Some electrons that were pulled out of the gas do not recombine, but are pulled away from the cathode, so appear to come from the cathode, (not the anode) |
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3.12.1.1 |
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01.5 |
Using magnetic fields The direction of deflection would show the charge Using electric fields The deflection would give an indication of mass |
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3.12.1.1 |
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02.1 |
The electrons emitted from the filament would collide with air molecules and not reach the screen. |
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3.12.1.2 |
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02.2 |
V1 is the p.d. in a circuit that is used to heat the cathode / cause thermionic emission To give the free/conduction electrons enough energy to escape V2 is the p.d. in a circuit that is used to accelerate the electrons |
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3.12.1.2 |
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02.3 |
\( \begin{array}{lll} eV & = & \frac{1}{2}mv^2\\ v & = & \sqrt{\frac{2eV}{m}}\\ & = & \sqrt{\frac{2 \times 1.61 \times 10^{-19} \times 4500}{9.11 \times 10^{-31}}}\\ & = & 3.98 \times 10^{7}\textrm{ m s}^{-1}\\ \end{array} \) |
Substitution Answer |
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3.12.1.2 |
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02.4 |
If the p.d. is decreased, the energy of the electrons is less When the electron hits the screen, it transfers less energy to the atoms in the phosphor/fluorescent screen Fewer electrons in the atoms of the screen are excited to a higher level and subsequently return to their original level Fewer photons are emitted from the screen, so the dot appears dimmer |
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3.12.1.2 |
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03.1 |
The largest specific charge known at the time was that of the hydrogen ion/proton The specific charge of the electron was larger So the electrons had a much bigger charge or a much smaller mass |
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3.12.1.3 |
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03.2 |
The magnetic force is downwards, and the current is from right to left By Fleming’s left hand rule, the magnetic field is into the page |
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3.12.1.3 |
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03.3 |
Increasing the accelerating p.d. increases the speed, and hence the magnetic force on the beam The accelerating p.d. needs to be increased to prevent the beam from being deflected. |
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3.12.1.3 |
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03.4 |
\( \begin{array}{l} \textrm{Magnetic force } = \textrm{ electric force so } Bev = Ee\textrm{ , so}\ Bv = E\\ E = \frac{V_{\rm{p}}}{d} = Bv,V_{\textrm{p}}\ = Bvd\\ eV = \frac{1}{2}mv^2,\,v = \sqrt{\frac{2e\ V_{\rm{acc}}}{m}}\\ V_{\rm{P}} = B \times \sqrt{\frac{2eV_{\rm{acc}}}{m}} \times d\\ V_{\rm{P}}^2 = {B^2}\frac{2eV_{\rm{acc}}}{m} \times {d^2}\\ {V_{{\rm{acc}}}} = \frac{m}{ed^2 B^2} \times V_{\rm{P}}^2\\ \textrm{A graph of }V_{\rm{acc}}\textrm{ against }V_{\rm{P}}^2\textrm{ has a gradient of }\frac{m}{ed^2}{B^2}\\ \end{array} \) |
Use of magnetic force and electric force Use of eV = \(\frac{1}{2}m{v^2}\) Equation of line |
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3.12.1.3 |
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03.5 |
\( \begin{array}{lll} \frac{e}{m} & = & \frac{d^2 B^2}{\rm{gradient}}\\ & = & \frac{0.150^2 \times 0.250^2}{830 \times 10^{-15}}\\ & = & 1.69 \times 10^{11}\textrm{ C kg}^{-1}\\ \end{array} \) Uncertainty = 0.35 × 10–15 kg C−1 m−2 T−2 \(\begin{array}{lll} \frac{e}{m} & = & \frac{d^2 B^2}{\rm{gradient}}\\ & = & \frac{0.15^{2} \times 0.25^{2}}{8.65 \times 10^{-15}}\\ & = & 1.63 \times 10^{11}\\ \end{array} \) Value = (1.69 ± 0.06) × 1011 C kg−1 |
Substitution Answer Alternate method: \(\textrm{% uncertainty } = \left( 0.35 \times \frac{100}{8.3} \right) = 4.2\%\)4.2% of 1.69 = 0.07 Answer with uncertainty |
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3.12.1.3 |
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04.1 |
The drag force acting vertically upwards is equal to the weight There is no net force, so no change of velocity/it moves at terminal velocity |
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3.12.1.4 |
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04.2 |
\( \begin{array}{l} \textrm{Weight } = mg = \frac{4}{3}\pi \rho r^{3}g\\ \textrm{Drag force } = 6\pi \eta rv\\ \frac{4}{3}\pi \rho r^{3} g = 6\pi \eta rv\\ \frac{4}{3}\rho {r^3}\ g = 6\eta v \\ r = \sqrt{\frac{9 \eta v}{2 \rho g}}\\ {\rm{Mass}} = \frac{4}{3}\pi \rho g \left( \sqrt{\frac{9\eta v}{\rho g}} \right)^3\\ \end{array} \) |
Use of drag force = weight Expression for r Expression for mass |
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3.12.1.4 |
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04.3 |
\( \begin{array}{l} mg = Eq = \frac{Vq}{d}\\ q = \frac{mgd}{V}\\ = \frac{3.2 \times 10^{-14} \textrm{ kg } \times 9.8 \times 12 \times 10^{-3}}{4700}\\ = 8.0 \times 10^{-19}\textrm{ C}\\ \textrm{Number of excess electrons } = \frac{8.0 \times 10^{-19}\textrm{ C}}{1.6 \times 10^{-19}\textrm{ C}} = 5\\ \end{array} \) |
Expression for e Answer Number |
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3.12.1.4 |
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04.4 |
The value of \(\frac{e}{m}\) is lower than it should be, so the mass is larger than it should be The value of the viscosity of air used was larger than it should be |
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3.12.1.4 |
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05.1 |
The negatively charged particles were not named electrons until later/until J.J. Thomson’s experiments |
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3.12.1.3 |
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05.2 |
The mark scheme gives some guidance as to what statements are expected to be seen in a 1 or 2 mark (L1), 3 or 4 mark (L2), and 5 or 6 mark (L3) answer
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The following statements are likely to be present: Bullet point 1 in question (Description of cathode rays in Crookes tube) 1. A tube is filled with low pressure gas 2. Gases at sufficiently low pressure in sealed glass tubes conduct electricity 3. The glowing gas near the anode, the positive column, was easily distorted when a magnet was brought near to the tube 4. This observation showed charged particles move through the gas when it conducts electricity 5. Radiation from the direction of the cathode forced ‘a paddle wheel’ placed in the tube to rotate. This radiation was referred to as ‘cathode rays’ 6. Neither the cathode nor the anode were heated Bullet point 2 in question (Description of cathode rays from thermionic emission) 7. When a cathode is heated, some of the electrons that move about freely inside the metal gain sufficient kinetic energy to leave the metal at its surface 8. The cathode is a wire filament, which is heated by passing an electric current through it 9. The filament or ‘cathode’ is at one end of an evacuated glass tube with a metal plate or ‘anode’ nearby 10. The electrons emitted from the filament are attracted to the anode by connecting a high voltage power supply between the anode and the cathode, with the anode positive relative to the filament 11. Because there are no gas molecules in the tube to stop the electrons, the electrons are accelerated to the anode where some of them pass through a small hole to form a narrow beam |
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3.12.1.1 |
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05.3 |
Crookes added a paddle wheel to the centre of the tube When the tube was working, the wheel moved because electrons/cathode rays were colliding with it |
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3.12.1.1 |
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05.4 |
Yes, the p.d. increases the energy according to eV = \(\frac{1}{2}\) mv2 |
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3.12.1.2 |
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06.1 |
Quarks have fractional charge |
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3.2.1.6 |
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06.2 |
\( \begin{array}{l} mg = Ee = \frac{Ve}{d}\\ V = \frac{mgd}{e}\textrm{, p.d. is proportional to }V\\ \textrm{if the charge is the same}\\ V = \frac{7 \times 10^{-5} \times\ g \times 3000} {3 \times 10^{-11} \times g}\\ = 7 \times 10^{9}\textrm{ V}\\ \end{array} \) |
Expression for V Assumption Answer |
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3.12.1.4 |
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06.3 |
For a force to be exerted, a charged object/particle needs to be moving |
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3.7.5.2 |
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06.4 |
The positrons are anti-electrons When they interact with electrons, they annihilate (producing two gamma rays), which reduces the charge on the sphere |
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3.2.1.3 |
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06.5 |
The resolution of his measuring instruments was not sufficient to distinguish between whole and fractional charge/there were no quarks produced on the oil drops |
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3.12.1.4 |
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07.1 |
Similarities: Both use electron beams In both methods, a force is exerted on the beam Differences: The force on the beam due to the magnetic field is perpendicular to the velocity The force on the beam due to the electric field is perpendicular to the plates |
Two similarities Two differences |
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3.12.1.3 |
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07.2 |
The vertical acceleration of the ball is 9.8 m s–2 The distance the ball falls is s = \(\frac{1}{2}\) at2 = \(\frac{1}{2}\) × 9.8 × t2 t = 0.319… s The time it takes the ball to travel 10 m is \(v = \frac{d}{t} = \frac{10}{0.319} = 31.3\ldots \) = 31 m s–1 |
Expression for time (explicit or implied) Use of equation of motion Answer |
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3.4.1.3 |
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07.3 |
The mass of the electron is extremely small, so the force of the Earth is negligible compared to the force of the magnetic or electric fields |
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3.12.1.3 |
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07.4 |
The force is perpendicular to the motion of the particles |
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3.7.5.2 |
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07.5 |
Force on an object moving in a circle \(\begin{array}{l} F = \frac{mv^2}{r} = Bqv\\ \textrm{So }\frac{mv}{r} = Bq\\ \textrm{And }v = \frac{2\pi r}{T} = 2 \pi rf\\ \frac{m2\pi rf}{r} = Bq\\ \textrm{So }f = \frac{qB}{2\pi m}\\ \end{array} \) |
Use of equation for circular motion and Lorentz force Substituting for v and T Answer |
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3.6.1.1 |
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08.1 |
Electrons in an atom in the screen move to a higher energy level when the accelerated electron transfers energy to the atom When the electron in the atom returns to its origin level, an X-ray is emitted The difference in energy between the levels must be sufficiently large to produce a photon with the wavelength/frequency of an X-ray |
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3.2.2.2 |
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08.2 |
Wavelength = 10−10 m \(\begin{array}{lll} \textrm{Energy } & = & \frac{{hc}}{\lambda }\\ & = & \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{10^{-10}}\\ & = & 1.99 \times 10^{-15}\textrm{ J allow }2 \times 10^{-15}\textrm{ or }10^{-15}\textrm{ J}\\ \end{array} \) |
Estimate Substitution Answer |
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3.2.1.3 |
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08.3 |
Energy = eV \(\begin{array}{lll} V & = & \frac{\rm{energy}}{e}\\ & = & \frac{1.99 \times 10^{-15}}{1.60 \times 10^{-19}}\\ & = & 12\ 400\textrm{ V}\\ \end{array} \) |
Substitution Answer |
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3.12.1.2 |
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08.4 |
A spark requires a very large potential difference, which can be produced by a transformer if there are very many more coils on the secondary than on the primary A sudden change in the p.d. across the primary produces rapidly changing flux in the core Which produces a very large p.d./induces a large e.m.f. as induced p.d. depends on the rate of change of flux and the number of turns |
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3.7.5.4 |
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08.5 |
\( \begin{array}{lll} \textrm{Electric field } & = & \frac{V}{d}\\ & = & \frac{10\,000}{0.01}\\ & = & 10^{6}\textrm{ V m}^{-1}\\ \end{array} \) |
Answer |
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3.7.3.2 |
Skills box Answers
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Question |
Answer |
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\( \begin{array}{l} eV = \frac{1}{2} mv^2\textrm{ is rearranged to obtain } v^2 = \frac{2eV}{m}\\ v = \sqrt{ \left( \frac{\left( 2 \times 1.6 \times 10^{-19}\textrm{ C } \times 30\textrm{ V}\right)} {\left( 9.11 \times 10^{-31}\ \rm{kg}\right)} \right)}\\ v = 3.2 \times 10^{6} \textrm{ m s}^{-1}\\ \end{array} \) (Check units: C V kg−1 = As V kg−1 = W V−1 s V kg−1 = J kg−1 = N m kg−1 = kg m s−2 m kg−1 = m2 s−2) |
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work done = \(\frac{1}{2}mv^2\) = \(\frac{1}{2}\) × 9.11 × 10−31 kg × (3.0 × 107 m s−1)2 = 4.1 × 10−16 J |
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work done = eV \(V = \frac{\textrm{work done}}{e} = \frac{1.2 \times 10^-{17}\ \rm{J}} {1.6 \times 10^{-19}\ \rm{C}} = 75\textrm{ V}\) |