Okay so the Rydberg Formula in the book is stated as
V= R{(1/(n1)^2)-(1/(n2)^2}
This makes sense to me, but in the example of practice problem 1.13, I had trouble understanding which N given would be N1 and which would be N2.
1.13 asks to calculate the wavelength of radiation generated by the transition from N=4 4o N=2.
I assumed that N=4 would be N1, because it is the initial condition. Therefore, N2 would be the N=2. However, the answer in the solutions manual has it the opposite way, with N=2 as N1 and N=4 as N2.
Could someone please help explain this? Many thanks.
Rydberg Formula, 1.13 HW
Moderators: Chem_Mod, Chem_Admin
-
- Posts: 31
- Joined: Fri Apr 06, 2018 11:05 am
Re: Rydberg Formula, 1.13 HW
Hello!
In order to find the change in energy level, the equation is E(final)-E(initial). Another way of thinking about this is if the temperature is 40 degrees F in the afternoon but it drops to 20 degrees F in the night, the change in temperature is -20 degrees F (change in temperature= final temp - initial temp). So in the example you gave, the initial energy level is N=4 and the final energy level is N=2. Therefore N1 is N=2 and N2 is N=4.
In order to find the change in energy level, the equation is E(final)-E(initial). Another way of thinking about this is if the temperature is 40 degrees F in the afternoon but it drops to 20 degrees F in the night, the change in temperature is -20 degrees F (change in temperature= final temp - initial temp). So in the example you gave, the initial energy level is N=4 and the final energy level is N=2. Therefore N1 is N=2 and N2 is N=4.
-
- Posts: 113
- Joined: Fri Apr 06, 2018 11:04 am
Re: Rydberg Formula, 1.13 HW
So for these types of questions, should we always assume that N1 stands for the final energy level and N2 is the initial energy level?
-
- Posts: 96
- Joined: Fri Apr 06, 2018 11:05 am
- Been upvoted: 1 time
-
- Posts: 23858
- Joined: Thu Aug 04, 2011 1:53 pm
- Has upvoted: 1253 times
Re: Rydberg Formula, 1.13 HW
The Rydberg formula always looks for a positive number (frequency). So, n2 > n1 always.
If you want to look in terms of energy, then you must always do E(final) - E(initial). But then when you carry this out, you are probably thinking about the energy of the electron. So, when energy is lost (negative value), this means that you had a positive value of energy being emitted due to conservation of energy. So, this positive value = h*freq, and you can solve for frequency.
If you want to look in terms of energy, then you must always do E(final) - E(initial). But then when you carry this out, you are probably thinking about the energy of the electron. So, when energy is lost (negative value), this means that you had a positive value of energy being emitted due to conservation of energy. So, this positive value = h*freq, and you can solve for frequency.
Return to “Bohr Frequency Condition, H-Atom , Atomic Spectroscopy”
Who is online
Users browsing this forum: No registered users and 5 guests