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.
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
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