Electron of energies 10.20 eV and 12.09 eV` can cause radiation to be emitted from hydrogen atoms . Calculate in each case, the principal quantum number of the orbit to which electron in the hydrogen atom is raised and the wavelength of the radiation emitted if it drops back to the ground state.

Answer

2 and 3

Explanation

We know the orbital energy of an electron revolving in `n^(th)` orbit is given by
`E_(n) = - (13.6)/(n^(2)) eV`
where `n` is the principal quantum number
When `n = 1, E_(1) = - 13.6 eV`,
`n = 2, E_(2) = - 3.4 eV`,
`n = 3, E_(3) = - 1.51 eV`.
energy needed by an electron to go from `L toL` level is `(13.6 - 3.4) = 10.2 eV` and that required to go form `K to M level is (13.6 - 1.51) = 12.09 eV`. The corresponding quantum numbers are `n = 2 to n = 3`, respestively. Hence electron will be raised to principal quantum numbers `2 and 3` corresponding to energies `10.20 eV and 12.09 eV`, respestively.
`n = 2 and 3`
When electron are coming back from `n = 2 to n = 3` to the ground state , i.e., `n = 1`. That is the case of Lyman series. Hence .
`(1)/(lambda_(1)) = R[(1)/(1^(2)) - (1)/(2^(2))] = (3 R)/(4)`
` implies lambda_(1) = (4)/(3 R) = (4)/(3 xx (10.97 xx 10^(6))) = 1216 Å`
and `(1)/(lambda_(2)) = R[(1)/(1^(2)) - (1)/(3^(2))] = (8 R)/(9)`
` lambda_(2) = (9)/(8 R) = (9)/(8 xx (10.97 xx 10^(6))) = 1020 Å`
`("Rounding off to nearest interger")`