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Biology and Chemistry Homework Help
Calculating the quantum state of an electron
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[QUOTE="bbbl67, post: 6081546, member: 587503"] [h2]Homework Statement [/h2] An electron in a hydrogen atom falling from an excited state to the ground state has the same wavelength than an electron moving at a speed of 7821 ms^-1. From which excited state did the electron fall from? [h2]Homework Equations[/h2] I used the kinetic energy equation: K = (m v^2)/2 m = 1 m_e = 9.109383×10^-31 kg v = 7821 m/s I also used the Rydberg Energy equation: E_n = -h c R_inf Z^2 / n^2 | E_n | electron energy h | Planck constant c | speed of light R_inf | Rydberg constant at infinity Z | atomic number n | quantum number Z = 1 (since it's hydrogen) n = quantum number = ? (this is what we are trying to find) The constant (h c R_inf) is also known as the Rydberg unit of energy, Ry: Ry = h c R_inf = 13.60569 eV [h2]The Attempt at a Solution[/h2] So this was in somebody's chemistry textbook, and I thought I'd try to solve this for giggles. I got no chemistry background, except in high school. I knew the kinetic energy equation by heart, I had to google the Rydberg equation ([URL]https://is.gd/QhBP9Z[/URL]). I wonder if my logic is correct here? kinetic energy: K = 2.786×10^-23 J = 1.739×10^-4 eV quantum energy: E_n = - Ry Z^2 / n^2 Since we're only looking for the magnitude in energy differences, I think we can ignore the negative sign in the equation, and simply write it as: E_n = Ry Z^2 / n^2 So we set, E_n = K K = Ry Z^2 / n^2 n^2 = Ry Z^2 / K n = sqrt(Ry Z^2 / K) = sqrt(13.60569 eV * 1^2 / 1.739×10^-4 eV) = 279.7 ~ 280 Conclusion: So that would mean that the electron fell from the 280th quantum state. How was my logic about this? Is there any other way of solving this? [/QUOTE]
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Calculating the quantum state of an electron
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