Singly ionized helium (He+) exhibits three distinct lines in its visible/near-IR
ID: 888200 • Letter: S
Question
Singly ionized helium (He+) exhibits three distinct lines in its visible/near-IR spectra with the wavelenghts 455 nm, 541 nm, and 1012 nm. These spectral lines are found usually in spectra of hot stars like Wolf-Rayet star and play an important role in solar astronomy. According to Bohr’s theory, these lines correspond to the set of transitions where the electron ‘jumps’ from a higher orbital to the n = 4 state.
(a) Find the quantum number of orbitals from which these transitions occur.
(b) Why transitions from two other orbitals to n=4 are missed? To answer this question, find the wavelength corresponding to these ‘missed’ transitions and compare with the spectral lines of hydrogen.
Explanation / Answer
a ) wavenumber = 1/wavelength = 109677[1/n1^2 - 1/n2^2] n2>n1
1. (1/(455*10^(-7))) = 109677((1/4^2) - (1/n2^2))
n2 = 2.7
2. (1/(541*10^(-7))) = 109677((1/4^2) - (1/n2^2))
n2 = 3
3. (1/(1012*10^(-7))) = 109677((1/4^2) - (1/n2^2))
n2 = 6
b) from 1,2 both were results n2 = 2.7,3 . which are lesser 4 .so that may not be
observed in spectrum.