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GR0177 #21
Problem
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Atomic$\Rightarrow$}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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duckduck_85
2008-10-24 22:57:53
Lyman: $1/\lambda=(1/1-1/4)R=3R/4$rnBalmer: $1/\lambda=(1/4-1/9)R=5R/36$rnrn$(1/Lyman)/(1/Balmer)=(4/3R)/(36/5R)=5/27$
dumbguy
2007-10-23 13:20:15
Why wouldn't we use the formula of 1/lambda=R(1/nf-1/ni)?
 jesford2008-04-03 19:38:12 This works too, but you forgot to square the nf and ni.
 s0crates2008-10-22 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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