GREPhysics.NET: GR0177 #21

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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Comments

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 $(1/Lyman)/(1/Balmer)=(4/3R)/(36/5R)=5/27$

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dumbguy
2007-10-23 13:20:15

Why wouldn't we use the formula of 1/lambda=R(1/nf-1/ni)?

jesford
2008-04-03 19:38:12

This works too, but you forgot to square the nf and ni.

s0crates
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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Lyman: $1/\lambda=(1/1-1/4)R=3R/4$ rnBalmer: $1/\lambda=(1/4-1/9)R=5R/36$ rnrn $(1/Lyman)/(1/Balmer)=(4/3R)/(36/5R)=5/27$

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