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How much of the atom's internal energy is released?

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  1. Oct 2, 2016 #1
    1. The problem statement, all variables and given/known data
    A uranium-238 atom can break up into a thorium-234 atom and a particle called an alpha particle, α-4. The numbers indicate the inertias of the atoms and the alpha particle in atomic mass units (1 amu = 1.66 × 10−27 kg). When an uranium atom initially at rest breaks up, the thorium atom is observed to recoil with an x component of velocity of -2.9 × 105 m/s.How much of the uranium atom's internal energy is released in the breakup?

    2. Relevant equations
    k=1/2mv^2

    3. The attempt at a solution
    i assumed that the energy released would come form the amount of kinetic energy of the thorium - the kinetic energy of the alpha particle. this got me and answer of -1.60x 10^-14. i believe that i hit the problem when calculating the mass to use of the atoms except i am unsure if this is even the correct way to approach the question
     
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  3. Oct 2, 2016 #2

    mfb

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    Both the kinetic energies come from the energy released in the decay. You have to sum the energies.
    How did you get the kinetic energy of the alpha particle?
    What did you use and where do you expect a problem?

    Your answer is missing units.
     
  4. Oct 2, 2016 #3
    i looked up the mass of an alpha particle and it was 6.64424. 10-27 kg. The i found the kinetic energy of that and it was 2.79390292x10^-16. next i found the mass of thorium by multipying 234 by 1.66 × 10−27 kg. Then i found the kinetic energy of thorium and add it to the alpha particle kinetic energy. the answer was 6.86x10^10 J. this is inncorrect. can you tell me where my error is
     
  5. Oct 2, 2016 #4

    haruspex

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    Please post all your working.
     
  6. Oct 3, 2016 #5
    ALPHA PARTICLE : mass= 6.64424. 10-27 kg(1.66 × 10−27 kg)=1.1029x10^-53 amu
    k=(0.5)(1.1029x10^-53)(-2.9 × 10^5)^2=4.63x10^-43 J
    THORIUM: 234(1.66 x10^-27)=3.88x10^-25 amu
    k=(0.5)(3.88x10^-25)(-2.9 × 10^5)^2=1.6315x10^-14 J
    1.6315x10^-14 J+4.63x10^-43 J=1.63x10^-15 J
     
  7. Oct 3, 2016 #6

    mfb

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    That is not right.
    You have to calculate the speed of the alpha particle first. It does not have the same speed as the thorium nucleus.
    That is wrong as well because you keep mixing units randomly.
     
  8. Oct 3, 2016 #7
    can you please explain more where i made my error instead of just saying its wrong
     
  9. Oct 3, 2016 #8
    is the mass of the thorium atom 234.0436 amu?
     
  10. Oct 3, 2016 #9

    I like Serena

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    Hey Emily! ;)

    Yep. So where you had amu before, it should be kg. Oh, and an alpha particle is 4.00 amu (2 protons and 2 neutrons).

    We can find the speed of the alpha particle from the total momentum. It must be zero since the Uranium atom was at rest.
     
  11. Oct 3, 2016 #10
    so am i able to use amu for mass?
     
  12. Oct 3, 2016 #11

    I like Serena

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    Yep. (Nod)
    It's a unit just like grams - just smaller, and tuned to atomic masses.
     
  13. Oct 3, 2016 #12
    would that make the kinetic energy of the thorium atom : k=(0.5)(234)(-2.9 × 10^5)^2=9.83x10^12J?
     
  14. Oct 3, 2016 #13

    I like Serena

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    That is a LOT of joules for such a microscopic particle. :oldsurprised:
    To get joules we have to convert the amu unit to kg. That is, multiply 234 amu by 1.66x10^-27 kg/amu.
     
  15. Oct 3, 2016 #14
    "THORIUM: 234(1.66 x10^-27)=3.88x10^-25 amu
    k=(0.5)(3.88x10^-25)(-2.9 × 10^5)^2=1.6315x10^-14 J"
    so this process from earlier was correct for thorium?
     
  16. Oct 3, 2016 #15

    I like Serena

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    The process was correct. The unit was wrong.
    It's 234 amu = 3.88×10-25 kg.
     
    Last edited: Oct 3, 2016
  17. Oct 3, 2016 #16
    but the amount of kinetic energy is still right?
     
  18. Oct 3, 2016 #17

    mfb

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    I said "use units" twice already. And not in the way of "I'll add some unit to the final result and hope that it is right." Use them consistently and you'll spot the majority of your errors on your own.

    Let's start from scratch with the masses:
    => Thorium-234 has a mass of 234 amu, and α-4 has a mass of 4 amu.
    You also know 1 amu = 1.66 × 10−27 kg

    How can you combine those two things to get the masses of the nuclei in kg?
    The kinetic energy for thorium is right, the one for the alpha particle is not.
     
  19. Oct 3, 2016 #18
    okay so the mass of the alpha particle is 4amu(1.66x10^-27kg)= 6.64x 10^-27 kg
     
  20. Oct 3, 2016 #19
    now i need to solve for speed of the alpha particle. can i do that by using mommetum ?
     
  21. Oct 3, 2016 #20
    so i used the equation 1/2mv^2=1/2mv^2
    √(1.61315x10^-14)/(0.5)(6.64x10^-27)
    v=2.2x10^6 m/s

    K=(0.5)(6.64x 10^-27)(2.2x10^6)^2
    =1.60x10^-14J

    sum: 1.60x10^-14J+1.63x10^-15 J
    =1.76x10^-14J

    Is this correct?
     
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