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# The Balmer series for the hydrogen atom corresponds to electronic transitions that terminate in the state with quantum number $n=2$ as shown in Figure $\mathrm{P} 42.10$ . Consider the photon of longest wavelength corresponding to a transition shown in the figure. Determine (a) its energy and (b) its wavelength. Consider the spectral line of shortest wavelength corresponding to a transition shown in thefigure. Find $(\mathrm{c})$ its photon energy and $(\mathrm{d})$ its wavelength. (e) What is the shortest possible wavelength in the Balmer series?

## a) $=1.89 \mathrm{eV}$b) $=658 \mathrm{nm}$c) $=3.40 \mathrm{eV}$d) $=365 \mathrm{nm}$e) $=365 \mathrm{nm}$

Quantum Physics

Atomic Physics

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### Video Transcript

for part of this question To calculate the energy of the photo on, we're gonna consider the longest wavelength, which has the initial state of in equals three. And then the final state we're told is N equals two. So they calculate this energy, we're gonna use the equation that says that the energy which we're gonna call E is equal to 13.6 electron volts, which comes from the ground state energy of the hydrogen atom multiplied by the difference between the, um, final stink, which is 1/2 squared minus the initial state 1/3 squared go. So carrying out this operation gives us our energy value of the photons plugging these values in. On carrying this out, we find this is equal to 1.89 electron volts. So now that we know the energy, we can calculate the wavelength. Since the energy and wave like the related by, uh, the equation e is equal to H C over Lambda. We know that lambda, Then the wavelength is equal to h C. Overeat. So Lamda is equal to H here Ages Plank's constant. We want to use the electron volt version of Plank's constant since our energy is an electron volts. Playing these values in C is also of course, the speed of light. We find that this is equal is 658 times 10 to the minus nine meters or 658 nano meters. You can use meters or nano meters. I'm gonna choose to use nano meters here, but you could also write it as 10 to the minus nine meters, which would be correct as well. For part C, it wants the photon energy which gives us the shortest wavelength in the transition from the figure. Okay, so the shortest wavelength again, we have the same equation. But this time it's coming from, uh, and equal to infinity. So the shortest wavelengths comes from in equals infinity. So we have 13.6 electron volts. Tell me you write that you have 1/2 squared start final status still in equals two that aren't initial state is infinity. So we're subtracting one over infinity and you could square infinity but infinity square to still infinity. So this is just one over infinity. But one over infinity is equal to zero. So this value here goes to zero. Carrying out this operation, we find that this is equal to 3.4 electron volts. We can box this and as the solution for C for D. It wants us to find the wavelength. So we're going again. Use the exact same equation as we did in part B to find the wavelength. So Lambda is equal to H C. Every begin H is in units of electron volts. Second so 4.1357 times 10 to the minus 15 electron bold seconds and see is still the speed of light in units of meters per second. Carrying out this operation, we find that this wavelength is equal to 365 times 10 to the minus nine meters or 365 nano meters. Wavelength is often given in units of nanometers, so it's convenient to write it in units of man o meters for easy comparison. And then lastly, Part E says, what is the shortest possible wavelength in the bomber? Siri's? Well, the bomber Siri's is going from some higher value down to n equals two. And so that means we found the shortest possible wavelength in the bomber Siri's. It's the wavelength that was given in part D. So the shortest possible wavelength of the bomber Syriza's Justin please 365 Nano meters go, Luckenbach said. It is our solution T.

University of Kansas

Quantum Physics

Atomic Physics