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# $\bullet$ A hydrogen atom initially in the ground state absorbs a photon, which excites it to the $n=4$ state. Determine the wavelength and frequency of the photon.

## 97 $\mathrm{nm}$

#### Topics

Electromagnetic Waves

Atomic Physics

### Discussion

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##### Christina K.

Rutgers, The State University of New Jersey

##### Farnaz M.

Simon Fraser University

### Video Transcript

okay. And this question, we're told we have a hydrogen atom and the ground stayed and goes to the ankles for ST. And then what is the wave length and frequency. So if it's going to the unequal wanna any before we need to find the energy difference. So that's gonna be given by this formula minus 13.6 times one over and final. So that's gonna be, um, on equals four and then you want to square about. It's like in general, this is gonna be one over and F squared where F stands for final and then won over initial score. This that's one over one. So basically, you want to take 13.6 and divided by 16 to get the to get the energy. So that's 0.85 TV. And then if we want to get the frequency we want to set this equal to each half so off, it's gonna be equal to point a five e v divided by age. So using H is equal to 4.136 Um, hi. It's 10 to the minus 15. I got 2.0, six times tend to the 14 hurts and we want to set this sequel. Oh, and then to get Lambda, we know that frequency is equal to see meters per second over Lambda. And so therefore, Lambda is equal Thio, see over F So we want to invert this frequency and multiply it by the speed of light. So if you 0.99 times 10 to the so Oh, I forgot to invert it. I got any answer that didn't really make sense. It's so that I got 123 I got 14. 59 um, nanometers. That seems to me that seems a bit two large. So I'm just gonna go reference the book. You can see if that matches up with some of the the Spectra. So I actually the figure 12 28.9 has the unequal for two unequal one. Oh, I see. And it looks like that should be a UV. That should be a UV photonic. This to me. This is not a you beef Otan. So there must be some error. Let me go back and look at the frequency calculation. So I got 0.85 and then I did 0.852 by two by each to point out five times. 10 of the 14. Me pause while I try to sort this out. Okay, I see the error that I made. I forgot to subtract this one. So this number is actually gonna be, um, 12.75. So just a calculator error. But it did cascade through. So let's go ahead and correct these 12.75 And then the frequency is now equal to 3.8 times 10 to 15. So go ahead and erase. That can actually leave this hero. So 3.0 eat times 10 to the 15. And then that puts the frequency, uh, 300 or Excuse me, um, 97.3 new meters, okay.

University of Washington

#### Topics

Electromagnetic Waves

Atomic Physics

##### Christina K.

Rutgers, The State University of New Jersey

##### Farnaz M.

Simon Fraser University