|- Hydrogen Like Ions - Singly Ionized Helium - Doubly Ionized Lithium and the 'une' Atom|
Singly ionized helium is an atom that has lost one of its electrons. We must expect its remaining electron to act like a hydrogen electron. There is only one difference between a hydrogen atom and a singly ionized helium atom; number of protons in the nucleus, so the nuclear charge. Thereby we have to modify our equation that we found for hydrogen atom. Only thing we have to do is to simply multiply the equation of the hydrogen atom by Z2 and we must get Z=2 since there are 2 protons in a helium atom. And when we arrange our formula for Z=2 we will get:
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.