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A bright violet line occurs at 435.8 nm in the emission spectrum of mercury vapor. What amount of energy, in joules, must be released by an electron in a mercury atom to produce a photon of this light?
$\mathrm{E}=4.56 \cdot 10^{-19}$
Chemistry 101
Chapter 6
Electronic Structure and Periodic Properties of Elements
Electronic Structure
Periodic Table properties
University of Kentucky
Brown University
University of Toronto
Lectures
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So what we're told for this problem is that in the emission spectrum for mercury, one of the lines produced has a wavelength of 435.8 centimeters were asked to do is to calculate the energy of the foe town that generates this line. Um, so do that. We've got I've got the two pretty standard light equations written here. What we're gonna do is actually combine them together. So we know that energy equals planks, constant times the speed of light divided by the wavelength. We're given the wavelength written here to the right and meters. So to solve this problem, it's a pretty straightforward calculation where we take the value of planks constant which is 6.626 times 10 to the negative 34th jewel seconds. We multiply that by the speed of light in meters per second. And then we divide that number by the given wavelength 4.358 times 10 to the negative seven meters. And the answer that will get out of that is 4.56 times 10 to the negative. 19th Jules
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