When solving this problem, I ran into a problem when trying to determine which n value should be used for E(final) and which should be solved for E(initial).
When I plugged in n=1 for E(final) and solved for n, I came up with a decimal energy level, which is not possible.
When I plugged in n=1 for E(initial), I got n=3, which is the correct answer.
Why do I need to plug in n=1 for E(initial) and not E(final) to obtain the correct answer to this question?
Atomic Spectra- Question 1.15 [ENDORSED]
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Re: Atomic Spectra- Question 1.15 [ENDORSED]
I solved this problem using the Rydberg equation instead of the bohr frequency condition.
You solved the n1=1 so for a line that's 102.6 nm you use the formula v=c/(lambda) = 2.922 x10^15 s^-1
Then use the Rydberg to solve for n2 which would give you n2=3
You solved the n1=1 so for a line that's 102.6 nm you use the formula v=c/(lambda) = 2.922 x10^15 s^-1
Then use the Rydberg to solve for n2 which would give you n2=3
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Re: Atomic Spectra- Question 1.15
In the question it states "In the ultraviolet spectrum of atomic hydrogen..."
The ultraviolet spectrum is the Lyman series, so n final=1.
The ultraviolet spectrum is the Lyman series, so n final=1.
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