| Date | May 2015 | Marks available | 8 | Reference code | 15M.2.HL.TZ1.7 |
| Level | Higher level | Paper | Paper 2 | Time zone | Time zone 1 |
| Command term | Calculate and Explain | Question number | 7 | Adapted from | N/A |
Question
This question is in two parts. Part 1 is about thermal properties of matter. Part 2 is about quantum physics.
Part 1 Thermal properties of matter
Part 2 Quantum physics
The diagram shows the end of an electron diffraction tube.
A pattern forms when diffracted electrons are incident on a fluorescent layer at the end of the tube.
Three ice cubes at a temperature of 0ºC are dropped into a container of water at a temperature of 22ºC. The mass of each ice cube is 25 g and the mass of the water is 330 g. The ice melts, so that the temperature of the water decreases. The thermal capacity of the container is negligible.
(i) The following data are available.
Specific latent heat of fusion of ice = 3.3×105 J kg–1
Specific heat capacity of water = 4.2×103 J kg–1 K–1
Calculate the final temperature of the water when all of the ice has melted. Assume that no thermal energy is exchanged between the water and the surroundings.
(ii) Explain how the first law of thermodynamics applies to the water when the ice cubes are dropped into it.
Explain how the pattern demonstrates that electrons have wave properties.
Electrons are accelerated to a speed of 3.6×107 ms−1 by the electric field.
(i) Calculate the de Broglie wavelength of the electrons.
(ii) The cathode and anode are 22 mm apart and the field is uniform.
The potential difference between the cathode and the anode is 3.7 kV.
Show that the acceleration of the electrons is approximately 3×1016ms−2.
State what can be deduced about an electron from the amplitude of its associated wavefunction.
An electron reaching the central bright spot on the fluorescent screen has a small uncertainty in its position. Outline what the Heisenberg uncertainty principle is able to predict about another property of this electron.
Markscheme
(i) use of M×4.2×103×∆θ;
ml = 75×10–3×3.3×105 / 24750 J;
recognition that melted ice warms and water cools to common final temperature;
3.4°C;
(ii) work done on water by dropping cubes / negligible work done;
W negative or unchanged;
water gives thermal energy to ice;
Q negative;
water cools to a lower temperature;
∆ U negative / U decreases;
bright and dark rings/circles / circular fringes;
maximum and minimum / constructive and destructive;
mention of interference / mention of superposition;
link to interference being characteristic of waves;
(i) (p=mev=) 3.28×10−23Ns;
\(\lambda = \left( {\frac{{\rm{h}}}{{\rm{p}}} = \frac{{6.63 \times {{10}^{ - 34}}}}{{3.28 \times {{10}^{ - 23}}}} = } \right)2.02 \times {10^{ - 11}}{\rm{m}}\);
(ii) \(E = \left( {\frac{{\Delta V}}{{\Delta x}}} \right) = \frac{{3.7 \times {{10}^3}}}{{22 \times {{10}^{ - 3}}}}\left( { = 1.68 \times {{10}^5}} \right){\rm{V}}{{\rm{m}}^{{\rm{ - 1}}}}\);
\(F = \left( {Eq} \right) = 1.68 \times {10^5} \times 1.6 \times {10^{ - 19}} = \left( {2.69 \times {{10}^{ - 14}}} \right){\rm{N}}\);
\(a = \frac{F}{m} = \left( {\frac{{2.69 \times {{10}^{ - 14}}}}{{9.11 \times {{10}^{ - 31}}}}} \right) = 2.95 \times {10^{16}}{\rm{m}}{{\rm{s}}^{ - 2}}\);
or
use of appropriate equation, eg v2 = u2 + 2as;
correct substitution (ignoring powers of ten);
a=2.95×1016 ms−2
square of amplitude (of wavefunction);
(proportional to) probability of finding an electron (at a particular point);
relates position to momentum (or velocity);
large uncertainty in momentum / most information on momentum is lost;
