Date | May 2012 | Marks available | 2 | Reference code | 12M.2.HL.TZ1.15 |
Level | Higher level | Paper | Paper 2 | Time zone | Time zone 1 |
Command term | Outline | Question number | 15 | Adapted from | N/A |
Question
Part 3 Atomic energy levels
Outline how atomic emission spectra provide evidence for the quantization of energy in atoms.
Consider an electron confined in a one-dimensional “box” of length L. The de Broglie waves associated with the electron are standing waves with wavelengths given by \(\frac{{2L}}{n}\), where n=1, 2, 3, …
Show that the energy En of the electron is given by
\[En = \frac{{{n^2}{h^2}}}{{8{m_e}{L^2}}}\]
where h is Planck’s constant and me is the mass of the electron.
An electron is confined in a “box” of length L=1.0×10–10m in the n=1 energy level. Its position as measured from one end of the box is (0.5±0.5)×10–10m. Determine
(i) the momentum of the electron.
(ii) the uncertainty in the momentum.
Markscheme
all particles have an associated wavelength / OWTTE;
wavelength is given by \(\lambda = \frac{h}{p}\), where h is Planck’s constant and p is momentum;
from de Broglie hypothesis, \({p_n} = \frac{h}{{{\lambda _n}}} = \frac{{nh}}{{2L}}\);
kinetic energy given by \({E_K} = \frac{{{p^2}}}{{2{m_e}}}\);
combined and manipulated to obtain result;
(i) \(\lambda = \frac{{2L}}{n} = \frac{{2 \times 1.0 \times {{10}^{ - 10}}}}{1} = 2.0 \times {10^{ - 10}}\);
\(p = \frac{h}{\lambda } = \frac{{6.6 \times {{10}^{ - 34}}}}{{2.0 \times {{10}^{ - 10}}}} = 3.3 \times {10^{ - 24}}{\rm{kgm}}{{\rm{s}}^{ - 1}}\);
Award [2] for alternative methods, e.g. calculating energy then momentum.
(ii) use of \(\Delta x\Delta p \ge \frac{h}{{4\pi }}\);
to get \(\Delta p \ge \frac{{6.6 \times {{10}^{ - 34}}}}{{4\pi \times 0.5 \times {{10}^{ - 10}}}} = 1.1 \times {10^{ - 24}}{\rm{kgm}}{{\rm{s}}^{ - 1}}\);