Ionization energy can be calculated using the equation given by Bohr that is shown below.
En=(−2.1799 aJ)Z2n2
Where,
Z is the atomic number of the atom.
n is the integer and the values are 1,2,3......
It is said that the electron fall to n=4 state. This means the final state nf=4. Emission line for Pickering series start from 5,6,7… and it ends at 4. This means the initial state will be ni=5,6,7.... and nf=4.
Wavenumber of any line that is present in visible spectrum of hydrogen is shown by Balmer and it can be given by using the formula shown below;
ν¯=RH(1nf2−1ni2)
In the above equation, RH is the Rydberg constant for hydrogen atom and it is equal to 1.097×107 m. The above equation can be written for the helium atom as shown below;
1λ=Z2RH(1nf2−1ni2) (1)
Wavelength of first five lines of Pickering series:
First line:
Atomic number of helium is 2. Photon is emitted when the electron moves from ni=5 to nf=5 state. Substituting this in equation (1), the wavelength can be calculated as shown below;
1λ=Z2RH(1nf2−1ni2)=4×1.097×107 m(142−152)=0.09873×107 mλ=10.09873×107 m=10.13×10−7 m=1013 nm
Second line:
Atomic number of helium is 2. Photon is emitted when the electron moves from ni=6 to nf=5 state. Substituting this in equation (1), the wavelength can be calculated as shown below;
1λ=Z2RH(1nf2−1ni2)=4×1.097×107 m(142−162)=0.1523×107 mλ=10.1523×107 m=6.566×10−7 m=656 nm
Third line:
Atomic number of helium is 2. Photon is emitted when the electron moves from ni=7 to nf=5 state. Substituting this in equation (1), the wavelength can be calculated as shown below;
1λ=Z2RH(1nf2−1ni2)=4×1.097×107 m(142−172)=0.1847×107 mλ=10.1847×107 m=5.4142×10−7 m=541 nm
Fourth line:
Atomic number of helium is 2. Photon is emitted when the electron moves from ni=8 to nf=5 state. Substituting this in equation (1), the wavelength can be calculated as shown below;
1λ=Z2RH(1nf2−1ni2)=4×1.097×107 m(142−182)=0.2057×107 mλ=10.2057×107 m=4.861×10−7 m=486 nm
Fifth line:
Atomic number of helium is 2. Photon is emitted when the electron moves from ni=9 to nf=5 state. Substituting this in equation (1), the wavelength can be calculated as shown below;
1λ=Z2RH(1nf2−1ni2)=4×1.097×107 m(142−192)=0.2203×107 mλ=10.2203×107 m=4.539×10−7 m=454 nm
The wavelength of the first five lines of Pickering series is summarized as shown below;
ni | 5 | 6 | 7 | 8 | 9 |
Wavelength (nm) | 1013 | 656 | 541 | 486 | 454 |
From the wavelength, it is found that this series lies in visible region of the electromagnetic spectrum.
In solar astronomy, this series is used n calculation of the properties of stars, age, luminosity, velocity, etc.