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Hydrogen atom energy

  1. Jan 21, 2012 #1
    1. The problem statement, all variables and given/known data

    In Bohr’s hydrogen atom, the frequency of radiation for transitions between adjacent orbits
    is νBohr = (En+1−En)/h. Classically, a charged particle moving in a circular orbit radiates
    at the frequency of the motion, νorbit = v/2πr. Find the ratio νBohr/νorbit for the Bohr
    orbits as a function of n. Evaluate the ratio for n = 1, 2, 5, 10, 100, and 1000, and thus show that νBohr/νorbit → 1 as n → ∞. This is an example of Bohr’s correspondence principle—that in the limit of large quantum numbers quantum-mechanical results agree with classical results.


    2. Relevant equations

    [tex] \nu_{Bohr}=(E_{n+1}-E_n)/h [/tex]

    [tex] \nu_{orbit}=\frac{v}{2 \pi r} [/tex]

    [tex] E_n= \frac{Z^2}{n^2}E_1 [/tex]

    3. The attempt at a solution

    From equation 3, can I say [itex] E_{n+1}= \frac{Z^2}{(n+1)^2}E_1 [/itex] ?

    Then doing the algebra gives me

    [tex] \frac{\nu_{Bohr}}{\nu_{Orbit}}=(\frac{1}{(n+1)^2}-\frac{1}{n^2} )\frac{Z^2E_1}{h} \frac{2 \pi r}{v} [/tex]

    Does that seem correct?

    However, the values from the brackets of [itex] \frac{1}{(n+1)^2}-\frac{1}{n^2} [/itex] gives me -3/4, -5/36, ... which don't tend to 1.
     
  2. jcsd
  3. Jan 21, 2012 #2

    Redbelly98

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    Looks right so far. But you have not accounted for the fact that r and v are not constant but depend on n, which will affect the value of the limit for large n. (For example, r is larger for higher n, since higher energy states correspond to larger orbits.)
     
  4. Jan 21, 2012 #3
    Now I have [itex] \frac{\nu_{Bohr}}{\nu_{Orbit}}=(\frac{n}{(n+1)^2}-1 )\frac{Z^2E_1}{h} \frac{2 \pi r}{v} [/itex] using [itex] r_n=\frac{n^2}{Z}a_0 [/itex].

    But even so, this only tells me that the brackets value is getting very small at large n.
     
  5. Jan 21, 2012 #4

    Redbelly98

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    You need to substitute an expression for vn as well.

    Also, this was probably just a typo on your part, but inside the brackets in your previous post it should be
    [tex]\left( \frac{n^2}{(n+1)^2}-1 \right)[/tex]
     
  6. Jan 21, 2012 #5
    I now have the ratio [itex] = (\frac {n^3}{(n+1)^2}-n)\frac{E_1 \hbar^2}{mk^2e^2} [/itex]

    But when I evaluate the non-bracket side, I get 1.28*10^-38.
    Multiplying this to the bracket doesn't give me a value that tends to 1.
     
  7. Jan 21, 2012 #6

    Redbelly98

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    Something is wrong, the units are not working out correctly.
    If you have replaced r with n2a0/Z, then r shouldn't be in your expression, and a0 should be there. Also, what expression for vn did you use? If you show more of your steps, we should be able to figure out where things went wrong.
     
  8. Jan 22, 2012 #7
    Is [itex] v_n=\frac{Zke^2}{n \hbar}=\frac{Z\alpha c}{n} [/itex]?
     
  9. Jan 22, 2012 #8
    Is this right? [itex] = (\frac {n^3}{(n+1)^2}-n)\frac{E_1 \hbar^2}{mk^2e^4} [/itex]

    The non bracket is now very big: 1.045*10^40.

    I think the bracket tends to -2.
     
  10. Jan 22, 2012 #9
    got it - thanks
     
  11. Jan 22, 2012 #10

    Redbelly98

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    Looks good.
    I get something different, perhaps you are not paying attention to units and just plugging in some numbers. So -- try including units in your calculation so that you can see them cancelling properly. If they don't cancel out, you probably have to convert some units somewhere. If it's still not working out after doing that, post your calculation of the non-bracket part here.
    I agree.
     
  12. Jan 22, 2012 #11

    Redbelly98

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    Ah, glad it worked out :smile:
     
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