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- Hydrogen Like Ions - Singly Ionized Helium - Doubly Ionized Lithium and the 'une' Atom |
So we can define the energy states for a singly ionized helium atom by using the formula above. When we give numbers for n, n=1, 2, 3, ..., we will get the energy value for each state. On the left, there is an energy diagram for a singly ionized helium atom. Now it is so easy to find the emission spectrum for a singly ionized helium.
Lets find the wavelength of the emitted photons when the single electron of the helium atom falls to the state n=3 from the state n=4, which corresponds to the first line of the Paschen series:
Energy of the system in the state n=4: | Energy of the system in the state n=3: |
Then we find the energy difference of these two states as 2.64 eV. From the energy difference we can find the wavelength of the photon as 470 nm.
This wavelength is in the visible part of the spectrum. Actually green. If the electron has fallen to n=1 from n=2 the incoming photon will have the wavelength; 30 nm. Which is in the ultraviolet region and very close to x-ray spectrum. Furthermore, if we want to ionize the remaining electron from the ground state, an external energy of 54.4 eV is needed.