Rydberg solving for an energy level
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- Опубліковано 9 вер 2014
- In this problem we are told the wavelength of a photon emitted from a hydrogen atom. We know the photon started in the n=5 energy level and we need to solve for what energy level it settled at.
Thank you very, very much for posting this video. It really helped me in my Chemistry Class. You explained it better than my teachers did and I really appreciate it!
Thank you SO much! I always got stuck on the algebra part of it for some reason and everyone just assumes you should know how to do it so nobody ever explains it... Your explanation helped me tons.
Thank you so so much. My TA did it completely wrong and I've been struggling to figure it out for several hours now. You're wonderful!
Thank you. For putting Paschem in mathematical sense. Once I saw the path I realized I was putting more into it than needed. :)
Thank you so much this is the first video I've come across that has actually helped me find the number of which energy level the electron moves to!!!
DUDE My teacher spent a whole month and I spent hours trying to understand this and you freaking helped in me in less than 10 minutes. Im sending this to my homies taking chem 111 lmfao
Not a dude, though. ;)
Love this! I wish you were my Chemistry 134 professor thank you sosososososo much
I REALLY needed this. Thank you so much!
This was super helpful, thank you!
THANK YOU , THANK YOU , THANK YOU, THANK YOU & THANK YOU SO VERY MUCH. YOUR EXPLAINATIONS REALLY HELPED ME OUT.
Thank you very much for the great explanation 👍👍👍🌹🌹
This was very helpful! Thanks 😁
I had to spam the like button real quick. I kept using the equation from my textbook which says that n1 squared is the final value and n2 squared is initial. I kept getting the wrong answer thinking it was me, but its actually my expensive textbook 🥴🥴. I did it this way and go the right answer. Thank you
same i was so confused LOL not sure if we in the same class but good luck on your exams!
Thank you! I have a final tomorrow, and I was sooo confused with how my professor was explaining it.
Thank you, that was very easy to understand.
thnks a lot for posting and explaining wonderfully
thank you so much!
Thanks so much!
Thanks a lot!!!
Bless your soul love Thankyou
I’m going to send this to my friends. Chemistry has been kicking our butts this semester lol
Thank you. OMG THANK YOU
Thank you so much
Thank you!!!!!!!!!!!!
thank you so much
THANK YOU :D
I keep getting 1.46 as my value for n no matter what I do, what am I doing wrong?
May God bless you
GOD BLESS
Thanks,
We were given E= -2.18*10^-18(1/nf^2 -1/ni^2) instead of 1/wavelength=1.097*10^7(1/nf^2-1/ni^2), so I was using that to solve for an energy level and could not figure it out. Using this equation instead finally gave me the right answer, but we were never taught that equation, I don’t know why.
SAMEE
she explains well but lol my ear burst when I used my earphones
Is this possible without a calculator for mcat?
Imoresive
6:58 I like how you said 9!
The scary thing is that I'm that enthusiastic and excitable in real life too....
Thanks
my chem teacher wants us to find n1 and n2 from wavelength, i cant figure it out
Thats literally impossible. Those numbers could be anything.
Okay this was 2 years ago but for anyone looking at this with the same problem yes it is impossible if this is all the information you have but with wavelength you can figure out one of the numbers because you can determine which series it's in. For example if it were 600 nm in wavelength you know that's in the visible spectrum and nf=2. Basically you will probably know one of the numbers it just won't be directly given to you
@@leilaloose5714 i have this same problem. "A UV photon found in the H spectra has a wavelength of 93.7 nm, from what “n” value did it relax?" do you know how i would find n1 and n2?
@@chloe-oy2ft so the question is asking you to find n1 I think but you know n2 because it's a UV photon so n2=1 (because all UV photons "jump down" to n=1) I'm not a teacher or anything so I don't know for sure and it's been a while since I've taken physics so definitely check my logic but I hope that helps :)
@@leilaloose5714 yes i ended up finding out the UV always jumps down to n=1, thank you sm! :)
How to find that initial state
I have a question Ma'am.. in my textbook they have given 1/n^2 initial - 1/n^2final as a part of the Rydberg equation. If I use this and say, I get a negative answer, and I use this to find out wavelength, then will not my answer be negative? How should I solve it instead?
Thank you Ma'am
You can’t have a negative wavelength. Think it through for a second. Wavelength is the length of a wave from peak to peak or trough to trough. It can’t be negative. That is why I don’t use initial and final for the n values. You have to set it up such that 1/n^2 - 1/n^2 gives you a positive number. Then you have to know if energy was released when the electron falls down or if energy was required to promote the electron.
When you converted nm to m, why is it (1m/1x10^9nm)? Isn't it supposed to be raised to the negative 9th power? (1m/1x10^-9)
See Myles Carter's comment below. In 1 meter there are 1 x 10 ^9 nm. There are a lot of nm in a m, so you need the positive exponent. The way you are thinking of it is 1 x 10 ^ -9 m = 1 nm. A nm is tiny, so it is only a small fraction of a m, hence the negative exponent in that second conversion factor.
I had a question that said "An excited hydrogen atom EMITS light with a FREQUENCY of 1.141 x 10^14 Hz to reach the ENERGY LEVEL for which N=4. In what PRINCIPAL QUANTUM LEVEL did the ELECTRON BEGIN?" The question seems very similar to the question you answered except they gave me a frequency rather then wavelength. I set the question up the same way except it was 1/lambda =1.097 x 10^7(1/16 - 1/n^2). When I solved the rest of the equation, I got the right answer. But in your equation the bracket was (1/n^2 - 1/25) which is the opposite direction. Im confused...I know you can't get a negative number, but is there some way Im supposed to tell which values go where?
+aiden Moore : Honestly, you have to select the values of n such that you don't get a negative number. It's not the best way to approach an equation, and you could memorize that n1 must always be < n2, but I hate memorizing things, so I just always try to make sure that the number can't be negative. For your example, n=4 must be less than n=?? because the electron 'fell down' to n=4. So, you must then put n=4 into the first spot because n1
Isn’t 1/.111158=8.996203602? Is it implied that we round up to set up the square root, for n has to equal a positive integer?
Yep. Energy levels are quantized. You can't have a partial energy level. So round the math to one sig fig and you get 9 which is a beautiful square of 3.
Taken to both extremes
Emit all energies = infinite black hole with an infinite number of singularities in a gamma field
Absorb all energies = a proton array with electron flow
God send
thank you! this really help :)
dear teacher, it appears that you plugged in the n=5 value into the opposite fraction. This gives the same value, but it should have a negative sign. Your voice and presentation are excellent though, after five videos I finally got it with your pleasant enthusiasm.
No... I didn't... The end result is an answer of n=3. 1/9 - 1/25 gives you a positive value... which is what you need to have. I make a big deal in the video about how (1/n^2 - 1/n^2) must give you a positive value. Thanks for the compliments, though. :)
🤗🤗🤗
what if the rydberg constant is in Joules (2.18x10^-18)
Then it is a totally different equation: Delta E = R (in J) x (1/n^2 - 1/n^2). The same stipulation applies that you must have the part in the parentheses equal to a positive number. Here, though, you aren't solving for wavelength of the photon emitted (or absorbed), but rather just the change in the energy of the electron going from one level to another.
Great explanation here: ch301.cm.utexas.edu/section2.php?target=atomic/H-atom/rydberg.html
thanks!
goat
I am not trying to doubt the teaching methods doc, but wouldn't it be more appropriate to teach by solving the equation for (n1)^2 first to give the equation of:
root([n2^2 (λ*R)]/[n2^2+(λ*R)])
Substituting n2, λ, and R to get the answer.
I am trying to teach my niece, and I was thinking doing algebra to derive these equations should be helpful for her later on. Do you recommend this approach?
Oh, sure, of course you could do it that way. I find that a lot of my students really struggle with trying to rearrange the equation first (for this type of problem) rather than plug and do algebra along the way. I definitely teach PV=nRT by rearranging the equation first because it is more straight forward than dealing with roots. Feel free to rearrange and then solve all you want. My students sometimes struggle with what 'taking something to the negative 1 power = the inverse of that something,' so I try to simplify things as much as I can.
Doc Lenczewski I remember, back when I was in high school they'd not give us full credit without showing the equation's derivation. I don't know what's going on with this generation. But thank you for the quick reply doc. Will be subscribed to your channel, I struggle when I teach, I don't know how you guys do it.
I think its much easier to first find energy of photon using E=hc/lambda. Set the energy of photon that you just solved for equal to -2.18E-18J(rydbergs constant). Divide the E-photon by rydbergs constant (-2.18E-18) making sure to use positive value (2.18E-18). then simply add 1/25 to the left side. You should now have 0.11119... on the left and still 1/nE2 on the right. Finally just invert the left side to 1/0.11119 and take square root of both sides. boom you get 2.99.... or 3
Yep, that is absolutely another way of solving the problem. It isn't using the Rydberg equation, it is using Bohr's work that led to new equation E=-2.18E-18J(1/n^2 - 1.n^2). It is almost identical to the Rydberg equation, it just uses energy instead of wavelength.
Savior
what if both initial and final n values are unknown?
Well, that's the more straight forward way then! You can plug in the n values (make sure to keep what is in parentheses equaling a positive value), and just solve for lambda.
So let me get this straight. Ive been struggling to understand the basic concept of this for two straight weeks and your telling me its as simple as plugging in the numbers? Can you come to my college please? I need to actaully understand this class to pass it and youre the only one that did it in under an hour.
How about you come to my college? ;)
how did U get 7.806*10^5m^-1??
+Durgesh Sakhardande : 1/1.281E-6=7.806E5 I just did one divided by 1.281 times 10 to the minus 6th.
Why can't you use E = hc/lambda to solve for E and then use a hydrogen energy level diagram to match the ∆E with two energy levels and thus get the answer?
Sure, you could absolutely do it that way... If you had a hydrogen energy level diagram happening to be lying about... Generally, though, I don't give one of those out during exams.
i see.. thank you for replying!
I'm cracking the fuck up, yo. Thank you SO, so much. Like. Yes. Like.
anyone know how to find the specific value of an energy level knowing wavelength and the two n values.
I'm not sure what you mean here by the 'specific value of an energy level'. The n values WOULD be the energy shell. You can convert wavelength into energy using E = hc/lambda (energy is Planks constant times the speed of light divided by wavelength in terms of meters). That energy would be the difference between the two shells.
@@doclenczewski727 but E = hf just gives U the energy of some photon of light that was emitted when a transition between energy levels took place. I need to construct an energy level diagram of sodium.
@@doclenczewski727 so I have the energy differences and I know what energy levels the transitions took place between eg n = 6 to n = 3. But now I need to calculate the actual energy value for n = 2,3,4 etc
@@doclenczewski727 I know the ground state is roughly -5.3 eV so now I need to find the n = 2 and so on but I don't know how
@@BenRankin I gotcha. There's a different equation for that. I don't typically ask my students for things in terms of eV, so I don't have it off the top of my head. It's something like E=-Z^2/n^2 * Rhc... My PChem text is up at school so I can't help much more than that right now. Good luck.
When I did 1 divided by 1.281 time 10 to the minus 6 I got 7.806 x 10^-7
Christina Pepi I did too 🤔
You two aren't doing order of operations correctly on your calculator. You have to be careful and ensure that the (times 10 to the minus 6) part is WITH the 1.281 part. Use parentheses if needed: 1/(1.281 x 10^-6)= and then you will get the correct value of 7.8125 x 10^5. What you are doing without the parentheses is 1/1.281 and then that answer x 10^-6. Another option is to use the EE button on your calculator. EE means (times 10 to the). So the math would be 1/1.281E-6. That gets rid of the necessity for the parentheses.
Hope that clears that up!
What if you're only given wavelength?
You can't solve for anything if you're just given the wavelength. If you look at the equation you have THREE unknowns: wavelength, n1 and n2. If you're only given one of the three, you can't just use one simple equation to solve for the other two. The only thing I can think is there might be an underlying assumption that the question has, like the electron is always in n=1 to start with and absorbs a photon with a certain energy. Then, in that case you would have two of the three unknowns: wavelength and n1, and you'd be solving for n2. If you want to write out the whole problem that you have, I'd be happy to help you through it, but just with the info you've given, you can't solve for anything using Rydberg if you just have the wavelength.
Doc Lenczewski I thought n2 was based on wavelength b/c visible wavelength usually has n=2
isn't nanometers 10^ -9 ?
+Myles Carter: Nanometers are smaller than meters. Around 2:30 min of the video, I convert nm to meters. In 1 meter there are 1 x10 ^9 nm. The way you are thinking of it is 1 x10^-9 m = 1 nm. Either way is fine, but you have to get a really small value when you convert nm to m because a lot of nm make up one meter.
+Myles Carter 1 nm is in fact 10^-9 but in this case we don´t have 1 we have 1281 so if you actually move the decimal point those 9 spaces you end with 0.000001281 m, now we have to get rid of all those 0s so just move the decimal point until you have 3 significant figures so 1.281x10^-6
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Annoying
Thank you so much