A student in a physics laboratory observes a hydrogen spectrum with a diffraction grating for the purpose of measuring the wavelengths of the emitted radiation. In the spectrum, she observes a yellow line and finds its wavelength to be 589 nm. (a) Assuming this is part of the Balmer series, determine $n_i$ , the principal quantum number of the initial state. (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?

Video Transcript

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 n f is 2 in this formula [30.13]. And we are gonna figure out what is n i, what was the initial energy level for this electron? So we have to solve for n i and we are gonna divide both sides by Rydberg's constant and that means 1 over n f squared minus 1 over n i squared is 1 over λ times R 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 n i squared after we switch the sides around is 1 over n f squared minus 1 over λR. 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 n 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 has to be an integer. So either the wavelength measurement is not correct or this is not a Balmer series emission and so n f is not 2 or this gas is not hydrogen at all.