Five line in the H atom spectrum have wavelength in A
A 1212.7
B 4340.5
C 4861.3
D 6562.8
E 10938
Three line result from transition to nf=2 visible series the other two result from transition in different series one with nf=1 and the other with nf=3 identify ni for each line
Use the Rydberg formula for this.
1/wavelength in meters = R(1/(nf)^2 - 1/(ni)^2 where R = 1.0974E7, nf is the final orbit and ni is the beginning orbit. You can look up R on the net for more places if you wish but if you use a calculator the value I show is good enough. First you want to identify the three lines where nf is 2. That's the Balmer series. Here is how you do one of them.
1/w = R(1/(nf)^2 - 1/(n2i)^2
1/w = 1.0974E7 (1/4 - 1/x^2). x can be 3, 4, 5, 6, etc.
Plug in 3 and solve for w. Check to see if that's one of the lines listed. Remember w will be in meters so you want to convert that to Angstroms when you compare with numbers from the problem.
1/w = 1.0974E7(1/4 - 1/9)
1/w = 1.0974E7(0.25 - 0.1111) = 1.0974E7(0.1389)
1/w = 1.524E6 and w = 6.561E-7 m = 6561 A and that matches the 6562.8 A shown. Remember you didn't use R for its exact value. Do that for ni = 4 and 5 and 6 or however many you need to identify the three lines from nf = 2. Having done that you want to find ni for the other lines. You do that using the same formula. Plug in w (remember to convert to meters) and solve for (ni)2 then for ni. Note: When you start this you don't know EITHER nf or ni but you can guess what to use at the start. You already know that you can't use 2 since you've done that for the Balmer series and those three lines you identified initally. So try 1 for nf = 1, then 2 3, 4, 5 etc for ni. That will identify one of the lines. The others one, now that you tried 1 and 2 for nf, would seem logical to try 3 for nf etc. I know this is long. It's a terribly long problem. Post your work if you get stuck.
Alright, let's break it down line by line, shall we?
Line A with wavelength 1212.7 Å is part of the Lyman series. Since we are given nf = 2 for this visible series, ni would be 1.
Line B with wavelength 4340.5 Å is also part of the Balmer series. But since this line corresponds to nf = 2, ni would be 4.
Line C with wavelength 4861.3 Å is another transition within the Balmer series, and it has nf = 2. So, ni for this line would be 3.
Line D with wavelength 6562.8 Å is a part of the Balmer series as well. Given that it is in the visible series with nf = 2, ni would be 3.
Line E with wavelength 10938 Å falls into the Paschen series. Since this line has nf = 2, ni would be 4.
So, to summarize:
Line A: ni = 1
Line B: ni = 4
Line C: ni = 3
Line D: ni = 3
Line E: ni = 4
Hope that clears things up! If you have any other questions, feel free to ask.
To identify ni for each line, we need to refer to the Balmer series of the hydrogen atom spectrum, which contains visible spectral lines. The Balmer series corresponds to transitions from higher energy levels (nf) to the n=2 energy level.
1. Line B with a wavelength of 4340.5 Å corresponds to the transition nf=3 to ni=2. (Transition from the third energy level to the second energy level)
2. Line C with a wavelength of 4861.3 Å corresponds to the transition nf=4 to ni=2. (Transition from the fourth energy level to the second energy level)
3. Line D with a wavelength of 6562.8 Å corresponds to the transition nf=∞ (continuum limit) to ni=2, as it is the first line in the Balmer series. (Transition from an energy level very far away to the second energy level)
The other two lines, A and E, belong to different series.
4. Line A with a wavelength of 1212.7 Å corresponds to nf=2 to ni=1. (Transition from the second energy level to the first energy level)
5. Line E with a wavelength of 10938 Å corresponds to nf=4 to ni=3. (Transition from the fourth energy level to the third energy level)
So, the identified ni for each line are as follows:
Line A: ni = 2
Line B: ni = 3
Line C: ni = 4
Line D: ni = 2
Line E: ni = 4
To identify the initial energy levels (ni) for each line in the hydrogen atom spectrum, we need to understand the series and transitions involved.
The Balmer series corresponds to transitions to nf = 2. So, three lines are from the Balmer series. The other two lines correspond to transitions in different series, one with nf = 1 and the other with nf = 3.
Let's match the wavelengths (in Å) with the corresponding series:
A - 1212.7 Å: This does not fall into any known series.
B - 4340.5 Å: This wavelength corresponds to the Hα line, which is the transition from nf = 3 to ni = 2 in the Balmer series.
C - 4861.3 Å: This wavelength corresponds to the Hβ line, which is the transition from nf = 4 to ni = 2 in the Balmer series.
D - 6562.8 Å: This wavelength corresponds to the Hα line, which is the transition from nf = 3 to ni = 2 in the Balmer series.
E - 10938 Å: This wavelength does not fall into any known series.
To summarize:
A - Unknown series
B - Balmer series: nf = 3, ni = 2
C - Balmer series: nf = 4, ni = 2
D - Balmer series: nf = 3, ni = 2
E - Unknown series