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Principal quantum number n, large ones for hydrogen

  1. Apr 12, 2009 #1
    hi folks, had a problem I couldn't do, though I made a valiant effort. perhaps you can tell me where I went wrong.

    1. The problem statement, all variables and given/known data
    All integer values of the principal quantum number n, even very large ones, are allowed in atoms. In practice, it is very hard to excite orbits that correspond to large n values in an atom unless the atom is totally isolated. Estimate the largest value of n that would be possible if you could make a gas of atomic hydrogen of density ρ = 7.1 10-10 g/cm3. For practical purposes, we regard an interatomic spacing of at least three times the diameter of the large-n atom as total isolation.
    n_max =

    2. Relevant equations
    r_n = a_0 * (n^2), Bohr radius equation
    a_0 = 5.3*10^(-11) m

    3. The attempt at a solution
    so I considered each atom as being in a cube L wide. L is the interatomic spacing, so L=3d=6*r_n=6*a_0*n^2=6*5.3*10^(-11)*n^2. our task is to find the value of n here.

    Since we're given a density, which equals Mass/Volume, I arbitrarily set Volume = 1 cm^3, to make things easier.
    Given a collection of these cubes stacked together in a large cube, say x*L is the side length. that is, there are x of these little cubes that each have a side length L. So this collection of cubes has a Volume of 1 cm3, therefore the side length has to be 1cm. so x*L=1cm=1*10^(-2)m.
    But we also know the density, given at 7.1*10^(-10) g/cm^3. Since we set V = 1, this means the mass = 7.1*10^(-10)g.
    The mass of the collection of little cubes is x^3 * m_e, the mass of an electron. (is this correct?...I just realized the mass of hydrogen, its proton, I completely ignored...). So I set x^3 = (7.1*10^(-10)g)/m_e, and substitute m_e = 9.109 * 10^(-28) g

    x^3 = (7.1*10^(-10)g)/(9.109*10^(-28)g) Solving for x, I get x=920300. this is the number of hydrogen atoms in our collection of little cubes.

    now, back to x*L=1*10^(-2)m. Solving for L, I get L = 1/(x*100) m. But since we know what L is, we can substitute it. so 6*5.3*10^(-11)*n^2=1/(x*100) m. We figured out x using the other route, so plugging x in, we get.

    6*5.3*10^(-11)*n^2 = 1/(920300*100) . now solving for n^2, I get

    n^2=1/(6*5.3*10^(-11)*920300*100), and

    n^2=34.1699, so that n=5.8455.

    I rounded this answer down to 5, since the the principal quantum number n has to be an integer, but this answer 5 was wrong.

    so. here are my thoughts. maybe i should just put 5.8455? or maybe I should rework the problem including the mass of hydrogen's proton? perhaps that is what led me to a wrong answer. i emailed the prof about integer vs. exact answer, and of course he never responded so... yeah.

    i'd appreciate any thoughts here or suggestions.

    Thanks much,
  2. jcsd
  3. Apr 12, 2009 #2
    turns out that both my concerns were valid. once I included the mass of the proton along with the electron, that led to the right answer. also, the prof/computer wanted an exact answer, so no rounding.
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