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Calculate the physical properties of a Bohr atom

  1. Oct 6, 2007 #1
    1. The problem statement, all variables and given/known data

    What is the radius of the n=1 orbit in C^5+ ? What is the energy of the electron in that orbit? What is the wavelength of the radiation emitted by C^5+ in the lyman alpha transition

    2. Relevant equations

    1/lambda=R*(1/(n^2)-1/(m^2) ); a(0)=(h/(2*pi))/(m*k*e^2);a(0)=.0529e-9m =5.29e-7 mE=-k*Z^2*e^2/2/((r)); r= a(0)*n^2/Z

    3. The attempt at a solution
    to find the radius , I know r=a(0)*n^2/Z ; Z is the atomic number. The atomic number of C^+5 is 5. r=(5.29e-7 m)*(1)^2/(5)= 1.06e-7 m

    to Find energy E=-k*Z^2*e^2/(2*r)= (9e9)((5)^2)((1.609e-19 J)^2)/(2*(1.06e-7m))=2.74e-20=.170 eV

    Now I'm asked to find the wavelength in the lyman alpha transition. Since I'm in the lyman transition , should I say one the transition state n=1? Maybe I don't need to find the other transition state. I could used the fact that 1/lambda=(1/hc)*(E(f)-E(i)) . I know from earlier in the problem that E(i) = .170 eV. For a lyman series E(f)= 13.6 eV . E(f)-E(i)/(hc)=1/lambda => lambda= hc/(E(f)-E(i))
     
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  3. Oct 6, 2007 #2

    learningphysics

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    The Lyman alpha transition is from m = 2... intial orbit, to n = 1... final orbit...

    so substitute these values into the formula and you'll get the wavelength.
     
  4. Oct 6, 2007 #3
    how do you know one of the transitions is m=2?
     
  5. Oct 6, 2007 #4

    learningphysics

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  6. Oct 6, 2007 #5
    so the rest of my calculations are okay? . USing the fact that E(f)-E(i)/(hc)=1/lambda is an option for finding lambda to right since E(f)=13.6 eV for the lyman series
     
  7. Oct 6, 2007 #6

    learningphysics

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    I'm not sure... what does C^5+ mean?
     
  8. Oct 6, 2007 #7
    C^5+ is Carbon and it reads that carbon has an atomic number Z +5. I finding some confusion here: according to the periodic table, the atomic number of carbon is 6.
     
    Last edited: Oct 6, 2007
  9. Oct 6, 2007 #8

    learningphysics

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    Does this problem involve the Bohr model of the atom? In that case En = -13.6ev*Z^2/n^2

    the energy at n = 1, comes out to: -13.6*5^2/1^2 = -340eV

    I think your calculation was wrong: it should be -(1/2)*k(5e)e/r (this is the potential energy + kinetic energy)

    and r =(5.29e-11 m)*(1)^2/(5)= 1.06e-11 m

    using these the energy also comes out to: -340eV

    I think the best way to get energy here is using En = -13.6ev*Z^2/n^2

    So yes, get the energy difference from n=2 to n =1... then set that equal to hc/lambda... then solve for lambda.
     
  10. Oct 6, 2007 #9
    Yes but doesn't E(0)=-13.6 eV only when we are talking about the hydrogen atom
     
  11. Oct 6, 2007 #10

    learningphysics

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    Ah... z = 6 not 5. sorry about that!

    I think the bohr model is for 1-electron atoms... I understand now that carbon 5+ refers to a carbon ion... a carbon atom that has lost 5 of its electrons, and just left with 1...

    like you found out carbon has atomic number 6. it lost 5 electrons, so it has just one left... So the bohr model still applies:

    En = -13.6ev*Z^2/n^2

    here if z = 1, then we're dealing with hydrogen... if we use Z = 6 then we have a carbon nucleus with 1 electron...

    E_1 = -13.6*6^2/1^2 = -489.6eV

    and using:

    -(1/2)*k(6e)e/r will also give -489.6eV

    so you can find E_2 etc...
     
    Last edited: Oct 6, 2007
  12. Oct 6, 2007 #11
    I still have two unknowns: E_2 and lambda. How would I find E_2? Why isn't E_1 -13.6 eV since the wavelength supposed to be in the lyman series

    I think I have it : E_1=E_0/n^2= 13.6 eV/1^2=13.6 eV. E_2=E_0/n^2=13.6/2^2=3.4 eV. Only problems I see with this line of reason is how do you know the other transition state is n=2?
     
    Last edited: Oct 6, 2007
  13. Oct 6, 2007 #12

    learningphysics

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    this is only for hydrogen... for other atoms:

    En = -13.6ev*Z^2/n^2

    so for z = 6, E1 = -489.6 eV... using the same formula E2 = -122.4eV

    I think that is just what "Lyman alpha transition" means.
     
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