Rydberg Equation
Вставка
- Опубліковано 12 лис 2012
- The Bohr model of the atom is used to calculate the wavelength of light emitted.
Warning! For some reason, this particular video gives a few watchers a lot of angst. Watch at your own risk, and let me clarify a few things. (1) I know that there are different ways to solve this problem. (2) I know that a different equation can be used (i.e., Balmer equation). (3) I know that particular forms of the Balmer equation may be used to preform the calculation more quickly.
Knowing all that, I still choose to solve this problem in this way. If you want more background on my solving method, see • Rydberg vs Balmer
I completely understand if you prefer to not watch this video for whatever reason. - Розваги
Amazing student-prof interaction.
Explains the Rydberg Equation really well. Thanks!
Thank you for breaking this down! I understood this better than using the Balmer’s formula.
Thank you for being clear. I'm here for conceptual reasons, but I will work a few problems after seeing this to get the idea in my brain. I actually can. It's not so bad. I really liked how you didn't puff your ego up by making this impossible to understand. We should have a book burning for impossible wording.
This is exactly what I needed. Thanks!
This was so helpful. Thank you!
thnk u so much for showing this. for that u deserve a sub
Hi! I'm from Turkey. I just find out about your videos while I was studying for my chem midterm which I have tomorrow. I didn't know anything but now I started learning wish I could heard about your videos long ago. Thank you SIR!
+Asli Kucukcalik I hope the exam went well!
7 yil sonra baska bir turk, ertesi gunku chem mt'sine calisiyor...
Very Helpful! Also, your voice reminds of of Amir from College Humor's Jake and Amir
Thanks for the video, very helpful!
It doesn't really matter as long as you know conceptually which way energy is flowing.
Thank you for helping us in lockdown
Love from kashmir (india)
The rydberg constant for hydrogen is: (1.097 x 10^7) I think. Where are you getting the: (2.179 x 10^-18) from?
i know your comment is 8 months old but i'll be answering it just in case you or someone else still has the same question...
he wrote the Rydberg formula which is ΔE=h · c · Rh (1/n - 1/n) as ΔE=Rh (1/n - 1/n)
@@shnig2333 Thanks! Bio student here wondering the same thing
Those who are wondering why rydberg constand have diff value so
Rydberg constant is 2π²k²Z²e⁴m / h³c
And this have value 1.097 × 10⁷
But he wrote rydberg formula= hc/wavelength
And according to that rydberg constant formula changes i.e.
2π²k²Z²e⁴m / h² which is energy formula
And equal to 13.6eV and if we convert it into Volt it will be 13.6 × 1.6×10-¹⁹ = 21.76× 10-¹⁹
This value shift according to orbit
Very helpful, thank you kindly
This was an exact question on my test.. must be the same problem bank..or just very coincidental
Different units - multiply yours by hc and it will be the same in mine (Joules).
Thanks dude!!
when you convert from nanometers to meters, why is it 10^9 and not 10^-9?
You can either say 1 nm = 10^-9 m or 1 m = 10^9 nm. Both are equivalent mathematically.
there are 1.0e^9 nm in one m
Liked it....💓
I think Rydberg's constant as that number some of you are typing represents a wavelength. If you multiply it by plank's constant and the speed of light you get the number as an electron volt (eV). From there you can convert it into Joules and it becomes 2.18*10^-18 J.
Someone should call me out if I'm wrong, because I'm not entirely sure.
It looks like the number you get when you multiply Rydberg's constant by Planck's constant and the speed of light is already in Joules. My Planck's constant was given in J*s though.
It's a matter of units.
Why this name is redberg constant????
Wow, I'm so glad you are still replying to questions here. I happen to be desperate to know how Balmer came up with his constant 364.50682. I understand that RH is 4/364.50682, but where does that constant of Balmer's come from? Can you offer any insight on that score?
+LRC Physics I don't have an answer to your question exactly, but I'll refer you to my other video which at least lists the other constants: ua-cam.com/video/DEhdJLUfd54/v-deo.html
If you find the answer to your question, please post here. Thanks!
Yeah, thanks. What I found out is nobody knows. It appears Balmer found his constant by trial and error. Then Rydberg divided 4 by it to get his, so the mystery is really the number 4, or 2 squared. It turns out that the hydrogen series follows the quadratic polynomial expansion n squared minus 1, which Rydberg inverted in his algebraic reformulation of Balmer's equation. It's all mysterious, but hidden, because Bohr's model explained it so easily.
@@lrcphysics8085 he is real legend
I have a question I hope someone can answer.
Since it is Delta E, shouldn't the formula be (1/nf^2 - 1/ni^2)*constant since the change should have final minus initial???
Thanks in advance! Very helpful video
Wait... I think I answered my own question. I'm using a negative constant where E= -hcR (1/n^2) where you use a positive constant hcR. Haha I'm just glad I figured it out. Thanks again!
Does the Rydberg Equation have to have the 1/initial - 1/final? I've looked around the internet and some have it 1/final - 1/initial and I don't understand why. If you reverse it does it make the equation 'absorption' specific or 'emission' specific or something?
I think the final answer has to be positive so which ever initial value is less than the final value - if you are emitting or absorbing it reverses the direction.
What happens if the initial orbit is larger than final orbit?
(for ex. ni = 253 and nf = 252...... "highly excited hydrogen atom")
Energy comes out to be negative, right? does negative energy make sense?
maybe it simply means that the hydrogen is losing energy, so the magnitude of the energy can be written regardless of the sign?
That's fine. Negative E means that E is being released. Positive E means it's being absorbed.
I'm not sure I follow this variation of the Rydberg equation. As I know it, the left side will be 1/lambda, not (delta)energy.
Inferring that 1/lambda = E should be fairly simple if true, but given that E = hc/lambda, and hc will never be equal to 1...
How does that work?
There are three variations that I'm aware of which use different R values. 1/lambda will need R=1.097x10^7 m^-1. ∆E uses R=2.178x10^-18 J. Finally, when the left side is frequency you use R=3.29x10^15 Hz or 1.097x10^7 m/s. Hope this helps!
I know the same equation :(
Hi, I'm from Peru, I can help with this question please. It reads: initially an electron is in the fourth level, if energy of 4.16 × 10 -7 erg is issued. At what level he moved? I get as data K = 3.29 x 10 ^ 15. Greetings.
Josue Ventura It's the same type of calculation. You are solving for nf knowing that ni is 4.
EnderlePhD ok thanks.
You can take value of hc together as 12400 but put wavelength in A° this would be easy for calculations
It's not necessary. The size of n1 in comparison to n2 tells you the direction of energy flow.
Hold up. Why is it n initial (6) - n final (2) in his equation? My chem book says final minus initial.
(1/6^2 - 1/2^2)
Vs
(1/2^2 - 1/6^2)
@T DP The two equations I have are
delta E = -2.18 x 10^-18 (1/n^2 final - 1/n^2 initial)
and
1/lambda = R(1/n^2 sub1 - 1/n^2 sub2) where n sub1 > n sub2 and R = 1.096776 x 10^7 m^-1
Thankyou!
sir will u plz explain when u solve ur equation there must be a negative value will come
I'm not sure I completely understand your question. In general, the negative shows the directions of energy flow.
Sir, does rydberg depends on atomic number
Rydberg's Formula does but it has to be changed (there's a missing Z^2), he has a video called Ionization Energy Calc 2 that I got here from that he explains that. I'm sad that Rydberg only works on Hydrogen-like atoms. I want to calculate every ionization energy. Edit: Any atom can be hydrogen-like, it just has to have only 1 electron left. So Helium's second ionization, Lithium's third ionization.
yea i thought the same thing. he is teaching a different way of finding the same answer.
Thank youuuuuu
In the version I was taught, the constant is just written negative ad the n values are switched. I was very confused for half of this video haha
great
How did you get 2.179*10^-18? Isn't Rydberg's constant=1.097*10^7?
That's for m^-1, convert it into joules by multiplying it by hc..... Because he wrote delta e
@@RohitKumar-hi4qf Omg thank you. I was so confused as I've been watching several videos and the constant is one or the other and I was not understanding why.
i don't understand how you find NI AND nF , somebody can explains ?
Ni stands for the initial orbit number and the Nf stands for the final orbit number
I like the part where he says to make the initial orbit larger just for fun but 12 is too large.
I'm not sure but I don't think 12 exists as far as orbits go. Just my amateur experience but I've only seen up to 8.
I thought the Rydberg constant was 1.097x10^7...?
I get it now
Why are you converting with 10^9 and not 10^-9? If anyone understands why please feel free to explain! Thank you!
+iTroNN Yes, thanks.
Thank you guys! I am just not used to ever seeing it done this way, for some reason we only ever have problems that require multiplying 10^-9 m / 1 nm. Now I can see that it is the same thing, it just took me a second. Have a great day everyone! P.S. Thank you for the video, it immensely helped me with my midterm!
+Ariel West Good luck on the rest of your classes!
Daaamnn!!
Double check your conversions and you'll see that 1 m = 1000000000 nm.
Wait but Rydberg's constant is -2.17*10^-18
Yes, that’s what was used in the video
I am here after 8 years 😅😅
2:20 sir ans is 4.85917*10^19j
Rydberg's constant is 109678 cm^-1 isn't it
It depends on how the equation is setup.
thanks for the help bill nye
Why does my chemistry course tell me RH = 3.29x10^15?
Aidan Ratnage It could be. There are different forms of the same equation. Check your equation carefully. Also, there are different series that cause the equation to vary as well.
+Aidan Ratnage That constant is for when you're solving the Rydberg equation for the frequency of the photon emitted, not the energy.
+MajorasMaskMailman Great.
@@EnderlePhD How do you know which one to use?
An easier and more correct way to calculate the wavelength would be to set the equation up like this: 1/λ=R(1/nf^2 - 1/ni^2).
Here it is worked out:
1/λ= 1.096 x 10^7 M^-1 (1/2^2-1/6^2)
1/λ= 1.096 x 10^7 M^-1 (8/36)
λ= 1/1.096x 10^7 M^-1
λ= 4.1058 x 10^7 M
λ= 410 nm
+Will Keegan Thanks, for the details of the calculation. I am very aware of that method. It turns out to be an equivalent method, not more or less correct. I use the method in this video for very particular reasons. If you desire to see why, go to: ua-cam.com/video/DEhdJLUfd54/v-deo.html
What if u want to calculate energy .....
With this u won't have to calculate lamda
O
Not much use full in india as iit exam ( world's 3rd most hard exam just requires the value as 109677 cm-1
Aditya Kanoi 3rd most? Source?
Life sucks and we're all going to die.
look at that shirt loose as hip Hop artist ...lol
You are pronouncing the name wrong wtf! its Rydberg like ride a bird.
What tha hell, no equations should be given for a test. Maybe constants, but that's about it.