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Ground state wavelength and uncertainty Bohr/deBroglie model

  1. Oct 1, 2015 #1
    A check my work question...

    The problem statement, all variables and given/known data


    Louis de Broglie tried to explain Bohr’s hydrogen atom electron orbits as being circles of just the
    right circumference such that an electron of the Bohr energy going around the circle will
    interfere constructively with itself. This seems to give a nice explanation of the angular momentum quantization.


    The attempt at a solution

    Q1: what would be the wavelength of such a wave in the ground state of hydrogen?

    ## \lambda n=2\pi r## from the Bohr energy equation we can tell n=1 gives the lowest energy state. We also have an equation for the possible values of r which reduces to ##r=a_{0}n^{2}## and so in our case ##\r=a_{0}##. Therefore ## \lambda n=2\pi r## becomes ## \lambda =2\pi a_{0} =0.332nm ##. Where ##a_{0}=0.0529##.

    Q2: According to the uncertainty principle, what would be the uncertainty in the wavelength, given that the electron position is known to be somewhere within the atom?

    At first I tried to do it using ##\Delta x \Delta p\geq \frac{\hbar}{2}##
    but then I found the expression ##\Delta k\Delta x\geq \frac{1}{2} ## in my text and so since uncertainty in position is equal to the radius of the atom (at least approximately) we have ##\Delta k \geq \frac{1}{2r_{atom}}## and therefore I just said the uncertainty in wavelength is just ##2r_{atom}##.

    Q3: Does this model make sense? Explain.

    Before I go trying to answer the last question and try to explain what I have done and found I'd like a little input on whether I am on the right track so far. Is my thinking correct here?

    Thank you so much!
     
  2. jcsd
  3. Oct 2, 2015 #2

    blue_leaf77

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    Science Advisor
    Homework Helper

    Q1. Eigenfunction of hydrogen atom is not an eigenfucntion of momentum, therefore there must be an uncertainty in finding a given momentum value. Since the de Broglie's hypothesis reads as ##|\mathbf{p}|=h/\lambda##, the quantity of interest in this case should be ##|\mathbf{p}|##, or equivalently the kinetic energy. Again, the electron momentum in hydrogen atom has some uncertainty, therefore we can't really find an exact value, rather it's more practical to calculate the average/expected value. So, I would suggest you calculate the expectation value of kinetic energy ##\langle E_k \rangle##, and take its square root to find the "expected" momentum ##|\mathbf{p}|##, from this point it should be straightforward to associate to the "expected" wavelength.
    PS: Words contained in double apostrophes "..." do not mathematically describe the actual averaged value.
    The strange thing with your result there is that, as you go farther from the nucleus (##n## becomes bigger), the wavelength decreases which means the electron moves faster. This is of course counter-intuitive as when you go farther you should feel less electrostatic force from the nucleus.
     
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