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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 P28.19. 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 the figure. Find (c) its photon energy and (d) its wavelength. (e) What is the shortest possible wavelength in the Balmer series?

## a) 1.889 $\mathrm{cV}$b) 658 $\mathrm{nm}$c) 3.02 $\mathrm{eV}$d) 412 $\mathrm{nm}$e) 366 $\mathrm{nm}$

Atomic Physics

Nuclear Physics

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##### Top Physics 103 Educators  ##### Christina K.

Rutgers, The State University of New Jersey ##### Aspen F.

University of Sheffield ##### Jared E.

University of Winnipeg

### Video Transcript

In this exercise, we have the bomber Siri's, which comes. It consists off transitions of the hydrogen atom between some initial energy level and why to a final one and F equals to two. Okay. And in the book, we have a figure that has some that shows some of the possible transitions in the bomber. Siri's in question A and B. We have to work with the the transition that results in the emitted photon have having the longest possible wavelength among those that are shown in the figure. So we'll have to consider. Is that first of all before doing any calculations? Ah, noticed that the energy of a photon is HC over Lunder. Okay. Ah, So if we're looking for the shortest, actually, in this case, the longest wavelength, we have to look for the the shortest energy for the smallest possible energy okay of the Fulton. But the energy of a photon that's emitted due to atomic transitions is equal to the energy of the in issue state miners, the energy of the final state. So it's e I minus cf. Okay, so this is just gonna call this dhoti. So if you're looking for the the smallest difference in energy possible. We can look up in the in the book and the figure and see which transition results in the smallest possible difference in energy. And from a figure, it's clear that the transition is it that results in the smallest energy is the traditions from N I equals 23 to the second, uh, the second state. Okay. In that case, the energy of the third state is given by minus one point 512 electoral votes that could be read from the figure and also from the figure have the energy of the second state. T two equals two minus 3.401 electoral votes. Okay, so the energy of the Fulton is equal to the three mine. Is it true? So this is miners. One went 5 12 plus 3.401 electoral votes, and this is equal to one point AIDS 89 electoral votes. That's the energy off the That's the the the energy of the longest of the photon that has the longest wavelength among those shown in the figure. And then we have to calculate the wavelength itself, and the wavelength is able to age See, over E hc is 1204 electoral votes in the meters E is 1.889 and it from votes. So Lunda is equal to 656 millimeters. Okay, now in question CND we have to look for the the transition that will result in the Fulton with the shortest wavelength among those in the fear. So again, uh, following a similar line of thought, as I did in question A for the for the wavelength Lambda Lambda to be maximum Ah, actually, here to be to be minimum we have we won the shortest wavelength, loved the possible. Then we want the maximum built A and among those in the figure, the maximum of lt happens in the transition between the sixth energy level and the 2nd 1 So the energy of the Fulton Belt is just gonna be ah SX minus two. And according to the figure, E six is minus 0.3 78 and e two is minus 3.401 So the difference in energy is even by 3.23 electoral votes In question D, we have to calculate the wavelength expensively and again. The wavelength is a silvery So this is go 140 and after in gold centimeters divided by the energy, which is 3.0 to 3 and it from votes. So this is 410 millimeters. Okay, in a question E, we have to calculate what is the shortest possible wavelength in the bomber. Siri's. Okay, um, and notice that, as I said before, the energy off the emitted electron is just dealt her you wish is the energy of the final. I'm sorry. The energy of the initial state and I e and E. I am sorry. Minus the energy of the second state. Because in the bomber series, the final state is always a second state. So this is equal to 13.6 times 1/4 minus one over, and I square. Okay, this In order to have this, you have to remember that the energy off the energy level off the hydrogen atom is minus 13.6 electoral votes, divided by and square. Um, and we're looking for the shortest wavelength possible. That means that we're looking for the greatest difference in energy and the greatest difference difference in energy. It happens when the second term this year zero It won over and I squared zero that iss when? And I tends to infinity. Okay, Well, when that happens, the energy no t is just 3.4 electoral votes and Lunda will be hc over that A and that's just 240 electoral votes and meters divided by 3.4 left in books. This is equal to three hundreds and 70. I'm sorry through other in 65 centimeters. Okay, this is the lower bound for the for the wavelengths of the of the bomber Siri's. Universidade de Sao Paulo

#### Topics

Atomic Physics

Nuclear Physics

##### Top Physics 103 Educators  ##### Christina K.

Rutgers, The State University of New Jersey ##### Aspen F.

University of Sheffield ##### Jared E.

University of Winnipeg