kevG Posted December 10, 2015 Report Share Posted December 10, 2015 So we had this question in class (Attached), but no one in our class knows how to do it as we cannot find the mark scheme. It would be great if anyone could tell us the answer with steps Thanks Reply Link to post Share on other sites More sharing options...
kw0573 Posted December 10, 2015 Report Share Posted December 10, 2015 (edited) There are two key components in this question1) Do peak intensity change?2) Does corresponding wavelength of the peak intensity change?Black body radiation demo: http://phet.colorado.edu/sims/blackbody-spectrum/blackbody-spectrum_en.htmlThe phenomenon is described through Wien's displacement lawwhich relates the peak shift and wavelength. At SL the relation is just presented matter-of-factly but I don't think HL delves too deep either because this law comes from Planck's Law (a vectored-valued function) so IB's "dumbing it down" for us.If you are looking for something for 1/16 T in the graph and you can't find anything, it's that the black body does not emit all wavelengths in equal proportions. The intensity predicted with Stefan-Boltzman * T^4 is a weighted average of all intensity by each wavelength. Intuitively, a cooler black body will emit lower energies (eg Earth emits infrared but Sun emits slightly higher-energy light, like visible and UV), and less intense (again, Earth and Sun) comparison. So I can't provide a satisfactory explanation since IB does not provide all the complex equations, but you must not confuse overall intensity with wavelength-specific intensity.EDIT:I looked it up, the Wien's displacement law is in HL syllabus. So as temperature halves, the peak frequency is doubled.The Stefan-Boltzmann Law helps to predict that the intensity will decrease. It seems that many sites, including the PHET demo above, shows the peak follow to T^4 proportionality, so that means the actual distribution of wavelengths is largely even. It could be that the book made a mistake but the general shape still points to B. Edited December 10, 2015 by kw0573 1 Reply Link to post Share on other sites More sharing options...
Vioh Posted December 10, 2015 Report Share Posted December 10, 2015 (edited) Like kw0573, I think the answer is B as well. Generally, with these types of multiple choice questions, you can just follow the process of elimination. First the question says that the area under the graph is halved, which means that C & D must be wrong. Now, as kw0573 already pointed out, you should use Wien's law to decide whether it's A or B. Wien's law states that Now after is halved, in order to keep the left-hand-side to still be a constant, must be doubled. Mathematically, So because originally, (or the peak) is at 0.1 mm, so the new peak should be 0.1 * 2 = 0.2 mm. Therefore, the answer is B. Edited December 10, 2015 by Vioh 1 Reply Link to post Share on other sites More sharing options...
kevG Posted December 10, 2015 Author Report Share Posted December 10, 2015 (edited) Thanks a lot guys, it all makes sense nowWe were confused between B and A Edited December 10, 2015 by gusmanK Reply Link to post Share on other sites More sharing options...
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