- $3.24$
- This is not an integer.
- Either the wavelength measurement is not correct, or this is not a Balmer Series emission, or the gas is not hydrogen.

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This is College Physics Answers with Shaun Dychko. A student observes an emission from a hydrogen atom of 589.0 nanometers and assumes that this is an emission from the Balmer series which means that <i>n f</i> is 2 in this formula [30.13]. And we are gonna figure out what is <i>n i</i>, what was the initial energy level for this electron? So we have to solve for <i>n i</i> and we are gonna divide both sides by Rydberg's constant and that means 1 over <i>n f</i> squared minus 1 over <i>n i</i> squared is 1 over <i>λ</i> times <i>R</i> and then we'll move this term to the right-hand side and move this term to the left-hand side and we are left with 1 over <i>n i</i> squared after we switch the sides around is 1 over <i>n f</i> squared minus 1 over <i>λR</i>. And then we can raise both sides to the exponent negative to flip this fraction and then make it one-half in order to take the square root so the exponent one-half is the same as writing a square root sign and the negative flips the fraction leaving us with <i>n i</i> on the left. And then what we do to the left, we also have to do to the right and so we wrap that in brackets and put exponent negative one-half and we'll leave it written that way. So the initial energy level then is 1 over 2 squared minus 1 over the observed wavelength— 589 times 10 to the minus 9 meters— times Rydberg's constant all to the power of negative one-half which is 3.24. And that is not reasonable because it's not an integer and <i>i</i> has to be an integer. So either the wavelength measurement is not correct or this is not a Balmer series emission and so <i>n f</i> is not 2 or this gas is not hydrogen at all.