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1. How to do this q? please.. i have been stuck on this for hours..

The “Humphreys series” is a less frequently discussed series in the atomic emission spectrum of hydrogen. The lowest energy line of the series is at 12371.9 nm and the sixth line of the series lies at 4376.5 nm. RH= 109677.8 cm-1

i)Find the principal quantum number of the upper and lower levels (n1,n2) for both of the above transitions.
2. What does the RH stand for?
3. (Original post by soph99jk)
How to do this q? please.. i have been stuck on this for hours..

The “Humphreys series” is a less frequently discussed series in the atomic emission spectrum of hydrogen. The lowest energy line of the series is at 12371.9 nm and the sixth line of the series lies at 4376.5 nm. RH= 109677.8 cm-1

i)Find the principal quantum number of the upper and lower levels (n1,n2) for both of the above transitions.
so atomic emissions of hydrogen are given by the Rydberg formula

E= RH [ (1/n1)2 - (1/n2)2 ]

the first line in the series will be from n2 = n1+1

Also 12371.9nm = 808.28cm-1

so putting the values into the Rydberg formula

808.28 = 109677.8 [ (1/n1)2 - (1/{n1+1})2 ]

This equation is undeniably a pain to solve.

However you now know you want to find the value of n1 such that

[ (1/n1)2 - (1/{n1+1})2 ] = 0.00737

hopefully you might know atleast the first three values of n1 correspond ot the Lyman, Balmer and Paschen series respectively, so your first try might be n1=4

1/16 -1/25 =0.0225
so not that

1/25 - 1/36 = 0.0122
so not that either

1/36 - 1/49 = 0.00737
Bingo

So the Humphreys series is transitions to n1=6

you can check this value for the other given transition.
4. (Original post by BDunlop)
What does the RH stand for?
Rydberg constant for Hydrogen.

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