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Problem 48 Medium Difficulty

Light of wavelength 121.6 $\mathrm{nm}$ is emitted by a hydrogen atom. What are the (a) higher quantum number and (b) lower
quantum number of the transition producing this emission? (c)
What is the name of the series that includes the transition?

Answer

(a) $n_{1}=2$
(b) $n_{2}=1$
(c) Referring to Fig. $39-18$, we see that this must be one of the Lyman series transitions.

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Top Physics 103 Educators
Elyse G.

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Video Transcript

in here we have a hydrogen atom which emits a wave. A light off the event 1 21.6 centimeters. So the hydrogen atom emits a light off a violent 1 21.69 meters. Let's evaluate the energy of this fortune. So the energy of the fort on in this case because really equal to let's see your Lambda. Since this isn't Nanometers will make use of the fact that it seems equal toe 12 40 electron golden tone on a meter divided by 1 21.6 millimeters. So this becomes equal toe, then point to electron worlds. Not this energy must have come from a certain transition. We make a transition from into to a certain anyone where this is the low quantum number and this is higher quantum number. The electron makes the transition it emits before done off a violent lambda. So in that case, he do minus even should be equal to the energy off the four done and we can write e to minus iguanas Negative off 13.6 Van wert into square minus for Noah en one square. No, it must be noted that such a high value off the ground. State energy will first notice that the downside energy it's always given us minus 13.6 Hueys now because we noted that such a high value of the Ford on energy must come from a transition. But the cab it men and women and who is very large and such high gaps are only seen for the few initial values like for anyone is able to 123 and four. If I make this small clearly, if this this one, then the difference between and one and two is very large. Then between 213 Reduces between three and four introduces even more between four and five produces even more, and then it keeps reducing. So let us try to put a value off two and one for this, when if it's absurd and two was equal to and end one as equal toe. One BC that this calculation exactly becomes 10 point build TVs. So in this case, our initial guests exactly solve Sansa so high. Oh, quantum numbers two and the loop quantum number is one, and because it makes a transition tow n equal to one, it's a lineman cities. So all the transitions, which ultimately land up at anyone equal to one, are always called Lyman Cities

University of Wisconsin - Madison
Top Physics 103 Educators
Elyse G.

Cornell University

Farnaz M.

Simon Fraser University

Jared E.

University of Winnipeg

Meghan M.

McMaster University