| Date | May 2015 | Marks available | 9 | Reference code | 15M.2.HL.TZ2.9 |
| Level | Higher level | Paper | Paper 2 | Time zone | Time zone 2 |
| Command term | Determine, Identify, and Sketch | Question number | 9 | Adapted from | N/A |
Question
This question is in two parts. Part 1 is about current electricity. Part 2 is about atoms.
Part 1 Current electricity
Part 2 Atoms
A 24Ω resistor is made from a conducting wire.
(i) The diameter of the wire is 0.30 mm and the wire has a resistivity of 1.7 × 10–8 Ω m. Calculate the length of the wire.
(ii) A potential difference of 12V is applied between the ends of the wire. Calculate the acceleration of a free electron in the wire.
(iii) Suggest why the average speed of the free electron does not keep increasing even though it is being accelerated.
An electric circuit consists of a supply connected to a 24Ω resistor in parallel with a variable resistor of resistance R. The supply has an emf of 12V and an internal resistance of 11Ω.
Power supplies deliver maximum power to an external circuit when the resistance of the external circuit equals the internal resistance of the power supply.
(i) Determine the value of R for this circuit at which maximum power is delivered to the external circuit.
(ii) Calculate the reading on the voltmeter for the value of R you determined in (b)(i).
(iii) Calculate the power dissipated in the 24Ω resistor when the maximum power is being delivered to the external circuit.
State what is meant by the wavefunction of an electron.
An electron is confined in a length of 2.0 \( \times \) 10–10 m.
(i) Determine the uncertainty in the momentum of the electron.
(ii) The electron has a momentum of 2.0 \( \times \) 10–23Ns. Determine the de Broglie wavelength of the electron.
(iii) On the axes, sketch the variation of the wavefunction \(\Psi \) of the electron in (d)(ii) with distance x. You may assume that \(\Psi \) \( = \) 0 when x \( = \) 0.
(iv) Identify the feature of your graph in (d)(iii) that gives the probability of finding the electron at a particular position and at a particular time.
Markscheme
99.7 (m);
Award [2] for bald correct answer.
Award [1 max] if area is incorrectly calculated, answer is 399 m if conversion to radius ignored, ie: allow ECF for second marking point if area is incorrect provided working clear.
(ii) electric field=\(\left( {\frac{{12}}{{99.7}} = } \right)0.120\left( {{\rm{V}}{{\rm{m}}^{ - 1}}} \right)\); (allow ECF from (a)(i))
electric force=(e×E=0.120×1.6×10-19 =)1.92×10-20(N);
acceleration\( = \left( {\frac{F}{m} = \frac{{1.92 \times {{10}^{ - 20}}}}{{9.1 \times {{10}^{ - 31}}}} = } \right)2.11 \times {10^{10}}(m{s^{ - 2}})\); } (5.27×109 if radius used in (a)(i) allow as ECF)
or
work done on electron = (Vq =)12×1.6×10-19 ;
energy gained by electron = me×a×distance travelled = 9.11 × 10-31 × a × 99.8;
2.11×1010 (ms-2);
Award [3] for a bald correct answer.
(iii) free electrons collide with ions and other electrons;
speed decreases during collisions / transfer their kinetic during collisions;
kinetic energy transferred to heat / wires have resistance;
and speed increases/acceleration until next collision;
(i) use of total resistance = 11Ω; (can be seen in second marking point)
\(\frac{1}{{11}} = \frac{1}{R} + \frac{1}{{24}}\);
20.3(Ω);
(ii) as current is same in resistor network and cell and resistance is same, half of emf must appear across resistor network;
6.0 (V);
or
\(I = \frac{{12}}{{\left( {11 + 11} \right)}} = 0.545\left( {\rm{A}} \right)\);
V=(0.545×11=) 6.0(V);
Other calculations are acceptable.
Award [2] for a bald correct answer.
(iii) pd across 2Ω=6.0V; (allow ECF from(b)(ii))
\(\left( {\frac{{{V^2}}}{R} = \frac{{36}}{{24}} = } \right)1.5\left( {\rm{W}} \right)\);
Award [2] for a bald correct answer.
measure of the probability of finding an electron (at a particular place and time);
(i) \(\Delta p = \frac{h}{{4\pi \Delta x}}\) and Δx=2.0×10-10; (both needed)
2.64×10–25(Ns); (also accept 5.28×10–25(Ns))
Award [2] for a bald correct answer.
(ii) \(\lambda = \frac{h}{p}\left( { = \frac{{6.63 \times {{10}^{ - 34}}}}{{2 \times {{10}^{ - 23}}}}} \right)\);
3.3×10-11 (m);
Award [2] for a bald correct answer.
(iii) 
periodic behaviour shown anywhere between 0 nm and 0.2 nm;
6 loops/repetitions shown anywhere between 0 nm and 0.2nm; } (allow ECF for division of 2×10-10 by answer to d(ii))
wavefunction completely fills from 0 nm to 0.2 nm and does not go beyond;
(iv) amplitude of Ψ/graph;
squared;
