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Balmer Series & Lyman Series

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Q: What are the widths of the wavelength intervals over which (a) the Lyman series and (b) the Balmer series extend? (Each width begins at the longest wavelength and ends at the series limit.) (c) What are the widths of the corresponding frequency intervals? Express the frequency intervals in tetrahertz (1THz = 10^12 Hz).

Dont understand what to do - and dont really understand Balmer and Lyman Series either - Any help with this question would be appreciated - thankyou!
 

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  • #2
ZapperZ
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ussrasu said:
Q: What are the widths of the wavelength intervals over which (a) the Lyman series and (b) the Balmer series extend? (Each width begins at the longest wavelength and ends at the series limit.) (c) What are the widths of the corresponding frequency intervals? Express the frequency intervals in tetrahertz (1THz = 10^12 Hz).
Dont understand what to do - and dont really understand Balmer and Lyman Series either - Any help with this question would be appreciated - thankyou!
Does your text not explain what these are, especially in the context of the Rydberg or hydrogen atom?

http://hyperphysics.phy-astr.gsu.edu/hbase/hyde.html#c2

Zz.
 
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Yeh but what i dont understand is how many different energy levels there are for each of the series? Like does it go on forever or does it stop at a certain number? Ie they start at n = 2 or n = 3, but where do they stop? Do they go up to infinity? Is it as simple as plugging the number in for each energy level into the formula and determining the wavelength?

Thanks
 
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SpaceTiger
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ussrasu said:
Yeh but what i dont understand is how many different energy levels there are for each of the series? Like does it go on forever or does it stop at a certain number?
In theory, they go to [itex]n=\infty[/itex]. In practice, this isn't the case, but you need not worry about such technicalities for this problem. Consider them to go to infinity.


Is it as simple as plugging the number in for each energy level into the formula and determining the wavelength?
Yup, though perhaps you should type out the formula you're using and tell us how you plan to use it to solve the problem.
 
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The formula is 1/lamba = R (1/2^2 - 1/n^2)

sorry if that doesnt make sense

But i will put in the number 2 in n for one wavelength and infinity for the other wavelength - and then subtract the 2 answers to get the wavelength range?

Is this along the right lines?
 
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SpaceTiger
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ussrasu said:
The formula is 1/lamba = R (1/2^2 - 1/n^2)
sorry if that doesnt make sense
But i will put in the number 2 in n for one wavelength and infinity for the other wavelength - and then subtract the 2 answers to get the wavelength range?
Is this along the right lines?
Looks right to me, at least for the Balmer series width.
 

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