# Thread: Frequency of absorption line in Hydrogen atom

1. # Of the following transitions in hydrogen atom, the one which gives an absorption line of highest frequency is:
a)n = 1 to n = 2
b)n = 3 to n =8
c)n = 2 to n = 1
d)n =8 to n =3

(c) and (d) cannot be true because it results in emission lines. Between (a) & (b), I think the answer is (a) because the energy difference between n = 1 to n = 2 is greater than the energy difference between n = 3 to n =8 levels and so higher is frequency of the absorption line. But the answer given in my book is (b). Is there a mistake in my reasoning?  2.

3. Okay, when in doubt use equations (Someone check me here though as im going off memory)

We know that f is proportional to 1/lambda which is proportional to (1/n<sub>1</sub>)<sup>2</sup> - (1/n<sub>2</sub>)<sup>2</sup> where n<sub>2</sub> is the starting energy level and n<sub>1</sub> is the ending energy level.

Subbing into this equation gives us that f<sub>a</sub>/f<sub>b</sub> = 7.85.. > 1 i.e. f<sub>a</sub> > f<sub>b</sub>

So at least we agree (Edit: Fixed Code typo - oops)  4. Originally Posted by river_rat
Okay, when in doubt use equations (Someone check me here though as im going off memory)

We know that f is proportional to 1/lambda which is proportional to (1/n<sub>1</sub>)<sup>2</sup> - (1/n<sub>2</sub>)<sup>2</sup> where n<sub>2</sub> is the starting energy level and n<sub>1</sub> is the ending energy level.

Subbing into this equation gives us that f<sub>a<sub>/f<sub>b</sub> = 7.85.. > 1 i.e. f<sub>a<sub> > f<sub>b<sub>

So at least we agree :-D Thanks for the guidance.  Bookmarks
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