Question
A wavelength of $4.653\textrm{ }\mu\textrm{m}$ is observed in a hydrogen spectrum for a transition that ends in the $n_f = 5$ level. What was $n_i$ for the initial level of the electron?
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Final Answer
7
Solution Video

OpenStax College Physics Solution, Chapter 30, Problem 19 (Problems & Exercises) (2:24)

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This is College Physics Answers with Shaun Dychko. A hydrogen atom emits a wavelength of 4.653 micrometers when an electron transitions from some initial state that we don't know to a final state of 5. Now equation [30.13] tells us what the wavelength is that's emitted as a function of the final and initial states of the electron and we need to solve this for <i>n i</i>. So let's first multiply through by Rydberg's constant and then get the <i>R</i> over <i>n i</i> squared term on the left side and that's gonna give us <i>R</i> over <i>n i</i> squared equals <i>R</i> over <i>n f</i> squared minus 1 over <i>λ</i>. And then we'll take the reciprocal of both sides but let's also get a common denominator here; we'll multiply top and bottom of this by <i>λ</i> and top of bottom of this by <i>n f</i> squared over <i>n f</i> squared and this makes <i>Rλ</i> minus <i>n f</i> squared all over <i>λn f</i> squared. And then we'll take the reciprocal of both sides which means flip these two fractions so that we get our unknown <i>n i</i> in the numerator. So it's <i>n i</i> squared at the moment divided by <i>R</i> and that equals <i>λn f</i> squared over <i>Rλ</i> minus <i>n f</i> squared. And then multiply both sides by <i>R</i> and then take the square root of both sides as well and we end up with <i>n f</i> to the power of 1 which because it's squared we can take it out of the square root sign and call it <i>n f</i> to the power of 1 times square root of <i>λ</i> times Rydberg's constant divided by <i>Rλ</i> minus <i>n f</i> squared. And so we have the final energy level we are told is 5 and we'll multiply that by the square root of 4.653 times 10 to the minus 6 meters times Rydberg's constant divided by the Rydberg's constant times the wavelength of the emitted photon minus <i>n f</i> squared so that's 5 squared and this works out to an initial energy level of 7.