A `12.5eV` electron beam is used to bombard gaseous hydrogen at room temperature. What serious of wavelength will be emitted?

Explanation

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.