It is given that the energy of the electron beam used to bombard gaseous hydrogen at room temperature is `12.5` eV. Also, the energy of the gaseous hydrogen in its ground state at room temperature is `-13.6` eV
When gaseous hydrogen is bombarded with an electron beam the energy of the gaseous hydrogen becomes `-13.6+12.5 eV i.e., -1.1 eV`
Orbital energy is related to orbit level (n) as:
`E=(-13.6)/((n)^(2))` eV
`E=(-13.6)/9=-1.5` eV
For n=3
This energy is approximately equal to the energy of gaseous hydrogen it can be concluded that the electron has jumped from n=1 to n=3 level During its de-excitation, the electrons can jump from n=3 to n=1 directly which forms a line of the Lyman series of the hydrogen spectrum We have the relation for wave number of Lyman series as:
`1/lambda=R_(y)(1/1^(2)-1/n^(2))`
Where,
`R_(y)` =Rydberg constant `=1.097xx10^(7)m^(-1)`
`lambda`=wavelength of radiation emitted by the transition of the electron
For n=3, we can obtain `lambda` as:
`1/lambda=1.097xx10^(7)(1/1^(2)-1/3^(2))`
`=1.097xx10^(7)(1-1/9)=1.097xx10^(7)xx8/9`
`lambda=9/(8xx1.097xx10^(7))=102.55nm`
If the electron jumps from n=2 to n=1 then the wavelength of the radiation is given as:
`1/lambda=1.097xx10^(7)(1/1^(2)-1/2^(2))`
`=1.097xx10^(7)(1-1/4)=1.097xx10^(7)xx3/4`
`lambda=4/(1.097xx10^(7)xx3)=121.54nm`
If the transition takes place from n=3 to n=2 then the wavelength of the radiation is given as:
`1/lambda=1.097xx10^(7)(1/2^(2)-1/3^(2))`
`=1.097xx10^(7)(1/4-1/9)=1.097xx10^(7)xx5/36`
`lambda=36/(5xx1.097xx10^(7))=656.33 nm`
This radiation corresponds to the Balmer series of the hydrogen spectrum Hence, in Lyman series two wavelengths i.e., `102.5` nm and `121.5` nm are emitted And in the Balmer series one wavelength i.e., `656.33` nm is emitted.
