| Date | May 2011 | Marks available | 1 | Reference code | 11M.3.SL.TZ1.21 |
| Level | Standard level | Paper | Paper 3 | Time zone | Time zone 1 |
| Command term | State | Question number | 21 | Adapted from | N/A |
Question
This question is about using a diffraction grating to view the emission spectrum of sodium.
Light from a sodium discharge tube is incident normally upon a diffraction grating having 8.00×105 lines per metre. The spectrum contains a double yellow line of wavelengths 589 nm and 590 nm.
Determine the angular separation of the two lines when viewed in the second order spectrum.
State why it is more difficult to observe the double yellow line when viewed in the first order spectrum.
Markscheme
\(d = \frac{1}{{8.00 \times {{10}^5}}} = 1.25 \times {10^{ - 6}}{\rm{m}}\);
\(d\sin \theta = n\lambda \Rightarrow \theta {\sin ^{ - 1}}\left[ {\frac{{n\lambda }}{d}} \right]\);
\({\sin ^{ - 1}}\left[ {\frac{{2 \times 589 \times {{10}^{ - 9}}}}{{1.25 \times {{10}^{ - 6}}}}} \right] = {\rm{ }}70.5^\circ {\rm{ }},{\sin ^{ - 1}}\left[ {\frac{{2 \times 590 \times {{10}^{ - 9}}}}{{1.25 \times {{10}^{ - 6}}}}} \right] = 70.7^\circ \);
70.7°−70.5°=0.2°;
the lines are closer together / not clearly separate in the first order spectrum;
