I'm having trouble figuring out to get started with this problem. Thanks in advance!
"A blue line is observed at 486.1 nm in the spectrum of atomic hydrogen. Determine the values of n for the beginning and ending energy levels of the electron during the emission of energy that leads to this spectral line."
n2=
n1=
Saping HW Q.8
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Re: Saping HW Q.8
I would convert the wavelength into meters and then find the frequency. Once you have the frequency, plug it into Rydberg's equation. Since it appears as a blue line, we know that it is part of the visible spectrum, so the Balmer series. That means that n1=2. From there you should be able to solve for n2, after inputting Rydberg's constant.
Hope this helps!
Hope this helps!
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Re: Saping HW Q.8
Ok so first off, you want to find the frequency of the electron using the equation c = λv. You should get that v = 6.17 x 1014 Hz. Then, you use Rydberg formula and say that 6.17 x 1014 = R((1/n22) - (1/n12)). Manipulate the equation so that you get n1 by itself on one side. Because the problem says that a blue line was emitted (or if you just look at the wavelength), you know that n2 is 2 (aka the visible region). After that, you should be able to solve for n1. You should get that n1 = 4, meaning that the electron fell from n = 4 to n = 2 and emitted a blue light as a result of the loss of energy
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Re: Saping HW Q.8
How do we know that n1=2? Is it characteristic of visible light in general or did it require solving?
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Re: Saping HW Q.8
Madison Wong 1G wrote:How do we know that n1=2? Is it characteristic of visible light in general or did it require solving?
it is based on the balmer series, which is the visible light and it starts at n=1
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Re: Saping HW Q.8
Would we be given a question where the initial n is not in the Balmar series? If so, do you think we'd be told what the initial n is?
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Re: Saping HW Q.8
I believe the only questions such as this one would have to be either in the Lyman series (n>1 to n=1) or Balmer series (n>2 to n=2), otherwise the question would be ambiguous.
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