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  GR0177 #21
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Atomic}Bohr Formula

One applies the Bohr formula E\propto (1/n_f^2 - 1/n_i^2) \propto 1/\lambda. For the Lyman radiation, this is E_L \propto (1-1/4). For the Balmer radiation, this is E_B \propto (1/4-1/9). Take the ratio of that to get E_L/E_B = \frac{3/4}{5/36}=27/5. But, since the problem wants the ratio of the wavelengths, which has an inverse relation to energy, one takes the inverse of that to get choice (B).

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2008-10-24 22:57:53
Lyman: 1/\lambda=(1/1-1/4)R=3R/4rnBalmer: 1/\lambda=(1/4-1/9)R=5R/36rnrn(1/Lyman)/(1/Balmer)=(4/3R)/(36/5R)=5/27NEC
2007-10-23 13:20:15
Why wouldn't we use the formula of 1/lambda=R(1/nf-1/ni)?
2008-04-03 19:38:12
This works too, but you forgot to square the nf and ni.
2008-10-22 15:12:04
That's essentially the same thing the original poster used, he just mentioned that the energy was proportional to 1/\lambda. Also, recall that Energy = \hbar*f and f = 1/\lambda, so that's where that relationship comes from.

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