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Problem 30 Easy Difficulty

Use Balmer's formula to calculate (a) the wavelength, (b) the frequency, and (c) the photon energy for the $H_\gamma$ line of the Balmer series for hydrogen.

Answer

(a) $$\mathrm{m}=434.1 \mathrm{nm}$$
(b) $$=6.906 \times 10^{14} \mathrm{Hz}$$
(c) $$\mathrm{J}=2.856 \mathrm{eV}$$

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Top Physics 103 Educators
Elyse G.

Cornell University

Marshall S.

University of Washington

Jared E.

University of Winnipeg

Meghan M.

McMaster University

Video Transcript

okay. In this problem, we're asked to look at the bomber at a line in the bomber. Siri's specifically that for the gamma lion and hydrogen and for that particular line, whereas to find the wavelength, frequency and energy of the photons corresponding to this line. So recall the equation. The bomber Siri's which gives us the wavelength. In terms of this, our value and a index for a lower and upper level R is one point zero nine seventy the negative seven inverse meters. So for the and recall for the H gamma, the age can align the lower value and is an equals two. The upper values in equals five and the text discusses the convention for how these air, determined on the upper level increases with the Greek letter. So H Alfa would be three ADA for Gammas five. That's how we get to these particular indices. So with that said, we can jump in, play these values in and find this lambda. So as I mentioned just playing some numbers in So for the lower we have won over the lower two squared. It's one of the upper five squared. All this is in first and After playing these numbers and playing in a value for our computing, we're left with a value of four hundred and thirty four And animators okay, we can relate this wavelength to frequency in an energy for photon using the expression e equals a Jeff also equal Nietzschean Landa Cool which tells us that f is equal to Siya Rhonda and he is a sea of Rwanda. Thus, we can find E to be a value that's computer using the speed of light and are given Lambda for thirty for even a nine. Get frequency first we have a frequency of six point nine one. I was tended. The fourteen hurts. We have an energy of using Planck's constant and the frequency we just found Very small energy four point five seven You negative nineteen meter and jewels so we can divide by one point six year the negative nineteen to convert TVs, which is probably more appropriate anyway. And we are left with two point eight six electron volts. Seems reasonable. Okay, these are our results into catching a couple these on the fly and that's it. It's all rest for it very much

University of Washington
Top Physics 103 Educators
Elyse G.

Cornell University

Marshall S.

University of Washington

Jared E.

University of Winnipeg

Meghan M.

McMaster University