.
Last time we used Bohr's postulates to derive the allowed energy levels and radii of hydgrogen, assuming the Balmer/Rydberg formula to help "guide" us.
We can do something far more impressive. Above, I used R_H taken from Balmers
empirical formula. But I can in fact derive (predict ) R_H! To do this,
we use the correspondence principle. This was Bohr's idea that, for
large quantum numbers, classical and quantum ideas should agree. Let's see what
this implies for hydrogen. Imagine dropping from level (n+1) to level n, where
n is very large. Then,
.
If n is large, then
,
which gives
Classically, we would expect that the frequency of emitted radiation would
equal the frequency of the orbiting electron. (The electron spins around the
atom with some orbital frequency, so it will emit light at that same frequency)
Classically, rotational frequency is
.
Now, recall that for a circular orbit,
Setting this classical frequency to the quantum frequency we found above (for large n) (this is the correspondence principle), we get
.
The correspondence principle has given us a_0 in terms of known constants, we
don't need to get R_H from experiment. In fact, we can now predict R_h,
from
(Exp't was 109,700 cm^-1. Fantastic agreement! In fact, the disagreement comes from the fact that the proton is not infinitely heavy, and can be easily corrected.)
Summarizing: Bohr's model, along with the correspondence principle, tells us
(1)
.
(2)
,
with R_H predicted:
,
and a_0 predicted
,
(3)
.
(To be honest, we haven't really derived this last formula. It was taken from
the empirical spectral results. But, you can show that the correspondence
principle requires that
)
The Balmer series becomes a complete, quantitative prediction. All the other series (with different n_i and n_f values) are also predictions. There's more! Bohr's theory now makes exact predictions for the spectrum of, e.g. He+, which is just like H, except that Z=2. So we predict for He+ that
,
and
.
Correspondence gives us
Putting this together:
This predicts some different wavelength from Hydrogen. In particular, note:
Some of the lines match up exactly! (Hydrogen Balmer 3 lines up with He 6->4, Hydrogen Balmer 4 lines up with He 8->4, etc.
Another prediction of this model is that, just like excited atoms can emit light, unexcited atoms can absorb light, lifting electrons up to higher levels. According to this picture, they will only absorb photons that excite exactly to an allowed energy.
Precisely these energy photons will be strongly absorbed, all others will pass through. Note that there are less possibilities available in the absorption direction. On the way down, all sorts of intermediate possibilities like n=3 -> n=2 are available.
This predicts (and of course is seen!)
(The details of these lines can teach you about e.g. density, temperature, and more, but we won't talk about this yet)
For hydrogen, you expect absorption of cold hydrogen to correspond to the Lyman series, where each line is dropping to n=1.
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