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Dino

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About Dino

  • Birthday March 31

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  • Favorite Area of Science
    Physics
  • Occupation
    Student

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  1. Dino

    Money

    I agree on the point that money is useful on how it is now, and I think it is the best to leave this topic as how it is, maybe in the future things might be diffrent, due to lack of resources. But that is the future, so let's leave this topic for now how it is.
  2. Dino

    Money

    But, let's say that 80% of the population would start making their own paintings then the shop owner would have so much options of paintings, he would only choose the best ones and eventually those who are not good enough would stop making paintings and they would work again to earn paintings. Please don't take this serious, just imagine.
  3. Dino

    Money

    What would happen if all money would dissapear and people only would be trading with for example art, so let's say a painting for grocery products for example, would the paintings become the same as money? Or is it then all based on opinions of a particular painting that determ the value?
  4. Would it be possible to create a material that does not absorb energy from the environment? So it can store energy in it self but it will not get more energy "extra". Maybe it would pass energy through the material but the material would not get warmer. I was just wondering this, I find it interesting to think about it.
  5. I thought that by travelling at the speed of light(or really close to it) you gain mass, and mass is a type of energy?
  6. There a thing I do not understand (actually a lott of things), a photon is a piece of energy traveling at the speed of light ? Does that mean that a photon keeps getting more energy because he is travelling at the speed of light ? [latex] E = mc^{2} [/latex] (Pleas correct me if I am saying anything wrong)
  7. So to eject a proton from a Fluorine nucleus there is a photon needed with 7,99485 MeV of energy going in the right direction to get absorbed by a proton of a Fluorine nucleus ? To calculate the wavelength of the photon there is this equation: or which brings up to a wavelenght of 1,550998455 * 10^-07 micrometer, multiplied by a 1000 to give nanometers it is 1,550998455 * 10^-04 nanometer which is 0,0001550998455 nanometer. Is this correct ? (Please correct me if I am saying anthing wrong)
  8. I am also not good in english but for example writing small texts or reading a lot of texts (for example a newspaper) improves your language skills. By seeing it multiple times, you will automatically remember the words and hopefully use them correctly in practice.
  9. So to eject a proton from a Fluorine nucleus there is a photon needed with 7,99485 MeV of energy going in the right direction to get absorbed by a proton of a Fluorine nucleus ? To calculate the wavelength of the photon there is this equation: [latex] {E}=\frac{\ hc}{\lambda } [/latex] or [latex] {\lambda}=\frac{\ hc}{E } [/latex] [latex] {\lambda(\mu m)}=\frac{\ 1,24}{7994850 } [/latex] which brings up to a wavelenght of 1,550998455 * 10^-07 micrometer, multiplied by a 1000 to give nanometers it is 1,550998455 * 10^-04 nanometer which is 0,0001550998455 nanometer. Is this correct ? (Please correct me if I am saying anthing wrong)
  10. So for example: to go from Fluorine to Oxygen-18 isotope, Fluorine = 18,99840 u 18,99840 * 931,494 = 17696,89561 MeV 17696,89561 - (9 * 0,511) = 17692,30 MeV Oxygen-18 isotope = 17,99916 u 17,99916 * 931,494 = 16766,11 MeV 16766,11 - (8 * 0,511) = 16762,02 MeV 17692,30 MeV + ? = 16762,02 MeV + 938,272 MeV 17692,30 MeV + 7,99 MeV = 16762,02 MeV + 938,272 MeV Are these calculations corrrect ?
  11. So for example: a Helium-4 isotope to a Hydrogen-3 isotope taking 1 proton away would take if my calculations are correct: Helium-4 has 2 protons and 2 neutrons, 2 * 1,007276466812 = 2,014552934 u and 2 * 1,00866491600 = 2,017329832 2,014552934 + 2,017329832 = 4,031882766 u Helium - 4 mass = 4,002603 4,031882766 - 4,002603 = 0,029279766 u 1 u = 931,494061 MeV 0,02927976 * 931,494061 = 27,27392814 MeV 27,27392814 / 4 = 6,818482034 MeV So does that mean that if I have calculated it correctly it takes 6,818482034 MeV to remove a proton of the nucleus from a Helium-4 isotope ?
  12. So does that mean that if technology is able to create a photon with sufficient energy and to let him move in the right direction to hit a proton we could change for example Helium into Hydrogen, this would be a great way to "change" all sorts of elements without the fear of extinction of them on this planet.
  13. So does that mean that if a photon with sufficient energy to eject a proton of a element that it would change into another element ? So for example: From Helium to Hydrogen ? Is this correct ?
  14. Dear reader of this topic, I was wondering what would happen if a photon gives his energy to a proton or a neutron in a atom. (Imagine that a photon would come across a single atom in space for example and he would miss the elektrons) Would the balance of the protons and neutrons disturb and what would happen next ? (Please correct me if I am saying anything wrong)
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