Jump to content


  • Content Count

  • Joined

  • Last visited

  • Days Won


Odelouche last won the day on May 5 2016

Odelouche had the most liked content!

Community Reputation

16 Good

Profile Information

  • Gender
  • Exams
    May 2016
  • Country

Recent Profile Visitors

The recent visitors block is disabled and is not being shown to other users.

  1. Happy to help Good luck for tomorrow!
  2. Hi, First of all, that function has to have a given domain and has to be continuous within that domain. Also, the integration of that function from the lower limit of the domain to the higher limit has to be equal to 1. Those are the conditions, now that you're given such a function, you need to find Q3 and Q1. To find Q1, you write the integration of the function from the lower limit of the domain to m, and set that integration equal to 0.25. Q1 = m. To find Q3, you do the same thing, only the integration is equal to 0.75 this time. Q3 should be bigger than Q1, and they should both be in the range of your domain (otherwise something has gone wrong). To calculate IQR, simply subtract Q1 from Q3. Hope this helps
  3. Hey guys, A couple comments about that exam: first of all, TZ0? From what I know, everyone in the world got the same paper, including South America. Is this normal when they change the syllabus? There were several parts that I couldn't do, but I'm that was because of lack of understanding or knowledge. However, there are a few legitimate doubts I have about some questions. First, when they gave you a nuclear reaction (can't remember the element, but the nucleon number was 32) and asked you to find the emitted energy? The mass of the daughter nucleus was larger than the mass of the reactant nucleus... Which implies an absorption of energy, not a release? Or did I miss something? Second, in the greenhouse effect question, for the second subpart, you had the amount of incident power on the outer atmosphere of Earth coming from the sun (1400W/m2), as well as the albedo of the Earth, 0.3 (albedo = scattered power/incident power). So you multiply 0.3*1400 to give you 420W/m2 of scattered power. Incident power - scattered power = absorbed power right? So 1400 - 420 = 980W/m2? But they wanted us to arrive at 245W/m2, which required us to divide 980 by 4, which I reluctantly did "for science" but without any understanding... Did I miss something, or are they wrong? Finally, every single person I've talked to about the HL exam said they had no idea how to do the last subpart of the exam, which required you to find the energy in Mev of an alpha particle emprisoned inside a gold nucleus, using Heisenberg's uncertainty principle. ???? First of all, in the uncertainty principle, energy is paired with time, which we do not have? Second, as its name suggests, the uncertainty principle deals with uncertainties, not actual values? And third, it's an inequality? Again, what am I missing here? I lost around 21 marks with absolute certainty, 30 marks with a conservative estimate of how mad it could go. The 7 is still in reach, praying for those mark bands to feel the force of gravity... Btw, did they just suddenly decide to remove all questions asking for definitions?? Or was that something they've announced? P.S: As I was writing this, I suddenly thought of something for this last subpart... We know the maximum allowed range of motion inside the nucleus of gold for the alpha particle (don't ask me how it got there) must be twice the radius of that nucleus (or the diameter), which was previously taken to be the least distance of approach, in the first part of the question. We use that as the uncertainty in position, considering the strong nuclear force and electromagnetic force must be bouncing that alpha particle everywhere inside that nucleus. Using Heisenberg's principle, we get an inequality for the uncertainty in momentum. Divide the whole thing by the mass of the alpha particle, you get it's minimum velocity, which you can then convert back to Mev through the equation for kinetic energy. It still only gives you an inequality, but maybe they're looking for the least possible energy of the alpha particle?
  4. Intuitive sense is all I look for in every question, and you've just given it to me! Thanks a lot for your explanation
  5. Hi, One way to answer your question is to consider dimensional analysis. You are looking for energy density, which is defined as the amount of energy contained in a unit volume of material or unit mass. Here, you need to use the unit mass definition. Hence, the unit is J/kg. You will also need the power output of your material, since it is intrinsically related to the amount of energy given off. You know that efficiency = output/input, so if you solve for power output, you get output = efficiency * input. Note that the efficiency should be in decimals, ie 0.35. The unit for power output is the watt, or J/s. Finally, you have mass. How do you combine J/s and kg to get J/kg? You divide the power output by the mass of material. This will give you (J/s)/(kg) = J/kgs. Notice that you have an extra unit of time in here, which doesn't seem to fit. Unless I did something wrong, that is because the data you have is flawed. Ideally, you need more than just the mass of material, you need how much mass of material is being used and for how long, ie kgs. To recap: 1. multiply input and efficiency to get power output. 2. Divide power output by mass used for x amount of time 3. Resultant is energy density. Another way to get the energy density if you only have mass and not mass*time is to use energy instead of power. For example if you have the energy input and efficiency, you do the same thing and you will not end up the extra time unit in your final answer. Hope this helps and good luck for tomorrow!
  6. Thanks a lot this makes a lot of sense! Small clarification: the fact that the currents aren't exactly superposed until a certain voltage is applied can be attributed to the fact that, as my teacher often says: "it's not a perfect situation"? Thanks!
  7. Hey, 'Nuclear' => nucleus, while 'atomic' => atom. The difference between those two are the electrons. So, the nuclear energy levels refer to the nucleus of the atom, and the arrangement and stability of neutrons and protons, while the atomic energy levels refer to positions and arrangement of electrons on their respective energy levels (ground state, first, second, etc...) according to the Pauli exclusion principle. In the question, it asks what provides evidence for the quantization of nuclear energy levels, thus referring to the protons and neutrons in the nucleus. In various types of decays, or in fusion and fission, alpha particles and photons are emitted, each with discrete energy levels. Therefore, I and II provide evidence. However, the atomic line emission spectra refers to which wavelengths of light are absorbed by certain elements, and this is in relation to the atomic energy levels (the photons of light are absorbed by the orbiting electrons, which consequently 'jump up' energy states around the nucleus). Therefore, the atomic line emission spectra does not provide evidence for the quantization of nuclear energy levels. Hope this helps
  8. Hi Guys, Any ideas for this one? Here is what I was able to figure out: since frequency is increased while intensity is constant, current flow must decrease. Hence, it's between A and C. That is because for both of those, the x-intercept is more to the left, meaning that it takes less voltage in order for the current flow to be 0 (makes sense because current flow is weaker when blue light is used). However, I'm stuck when it comes to differentiating between A and C. If anyone could tell me if my reasoning so far is correct, and what are the final steps, that would be greatly appreciated, Good luck tomorrow for everyone! P.S: the answer is C.
  9. Hi Guys, I'm having a hard time understanding this question: I failed it miserably in my mock, then went back, studied the concepts, tried again, and again couldn't come up with the right answers. I don't really understand what the mark scheme is doing, an explanation of the equations used as well as how the direction of current is determined and used would be much appreciated. The question and mark scheme are attached, thanks a lot for your replies!
  • Create New...