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Can anyone help with this Physical chem problem

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.


ii) Predict the wavelengths of the lines in this series that lie between 12371.9 nm and 4376.5 nm.


iii) What is the ionization energy, in eV, for the H atom from the orbital that is the lowest level of
this series?



Been trying to visualize this for ages, but getting nowhere - can anyone shed any light on any of these questions?!
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Avatar for BJack
BJack
19
The emission spectrum comes from excited electrons falling from various higher-energy orbitals to a particular lower-energy one. The lowest energy transition will therefore be something like n=3 -> n=2 or n=12 -> n=11. You should be able to use the Rydberg formula to work out what the relevant energy levels are for the given lines and use those to fill in the gap.

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