| Date | May 2011 | Marks available | 2 | Reference code | 11M.2.HL.TZ1.10 |
| Level | Higher level | Paper | Paper 2 | Time zone | Time zone 1 |
| Command term | Calculate | Question number | 10 | Adapted from | N/A |
Question
This question is in two parts. Part 1 is about quantum aspects of the electron. Part 2 is about electric circuits.
Part 1 Quantum aspects of the electron
The wavefunction ψ for an electron confined to move within a “box” of linear size L =1.0×10−10m, is a standing wave as shown.
State what is meant by a wavefunction.
State the position near which this electron is most likely to be found.
Calculate the momentum of the electron.
The energy, in joules, of the electron in a hydrogen atom, is given by \(E = \frac{{2.18 \times {{10}^{ - 18}}}}{{{n^2}}}\) where n is a positive integer. Calculate the wavelength of the photon emitted in a transition from the first excited state of hydrogen to the ground state.
The electron stays in the first excited state of hydrogen for a time of approximately Δt=1.0×10−10s.
(i) Determine the uncertainty in the energy of the electron in the first excited state.
(ii) Suggest, with reference to your answer to (e)(i), why the photons emitted in transitions from the first excited state of hydrogen to the ground state will, in fact, have a small range of wavelengths.
Markscheme
a function whose (absolute squared) value may be used to calculate the probability of finding a particle near a given position / quantity related to the probability of finding an electron near a given position/at a given position;
middle of the box / (near) 0.5×10−10m;
the de Broglie wavelength is 2.0×10−10m;
\(p = \frac{h}{\lambda } = \frac{{6.63 \times {{10}^{ - 34}}}}{{2.0 \times {{10}^{ - 10}}}} = 3.3 \times {10^{ - 24}}{\rm{Ns}}\);
difference in energy is
\(\Delta E\left( { = - \frac{{2.18 \times {{10}^{ - 18}}}}{{{2^2}}} + \frac{{2.18 \times {{10}^{ - 18}}}}{{{1^2}}}} \right) = 1.635 \times {10^{ - 18}}{\rm{J}}\);
\(\lambda = \frac{{hc}}{{\Delta E}}\);
\(\lambda = \left( {\frac{{6.63 \times {{10}^{ - 34}} \times 3.0 \times {{10}^8}}}{{1.635 \times {{10}^{ - 18}}}}} \right) = 1.22 \times {10^{ - 7}}{\rm{m}}\);
(i) attempt at using the energy – time uncertainty relation;
\(\Delta E\left( { = \frac{h}{{4\pi \Delta t}} = \frac{{6.63 \times {{10}^{ - 34}}}}{{4\pi \times 1.0 \times {{10}^{ - 10}}}}} \right) = 5.3 \times {10^{ - 25}}{\rm{J}}\);
(ii) the wavelength of the photons is determined by the difference in energy between the two levels;
and that energy difference is not well defined/definite/not always the same (because of the uncertainty principle);
