Electrons of energy `12.09 eV` can excite hydrogen atoms . To which orbit is the electron in the hydrogen atom raisd and what are the wavelengths of the radiations emitted as it dropes 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")`