A `100 eV` electron collides with a stationary helium ion `(He^(+))` in its ground state and excites to a higher level. After the collision , `He^(+)` ion emits two photons in succession with wavelength `1085 Å and 304 Å`. Find the principal quantum number of the excite in its ground state and. Also calculate energy of the electron after the collision. Given `h = 6.63 xx 10^(-34) J s`.

Answer

`5, 47.7 eV`

Explanation

The energy of the electron in the `n^(th)` state of `He^(+)` ion if atomic number `Z` is given by `E_(n) = - (13.6 eV) (Z^(2))/(n^(2))`
For `He^(+)` ion, `Z = 2`, therefore
`E_(n) = - ((13.6 eV) xx (2)^(2))/(n^(2))`
` = - (54.4)/(n^(2)) eV`
The energies `E_(1)` and `E_(2)` of the two emitted photons , in `eV`, are
`E_(1) = (12431)/(1085) eV = 11.4 eV`
and `E_(2) = (12431)/(304) eV = 40.9 eV`
Thus, total energy `E = E_(1) + E_(2) = 11.4 + 40.9 = 52.3 eV` Let `n` be the principal quantum number of the excited state. Using Eq. (i) , we have for the transition from `n = n to n = 1`.
`E = - (54.4 eV) ((1)/(l^(2)) - (1)/(n^(2)))`
But `E = 52.3 eV`, therefore
`52.3 eV = 54.4 eV xx (1 - (1)/(n^(2)))`
or `1 - (1)/(n^(2)) = (52.3)/(54.4) = 0.96`
`implies n^(2) = 25 or n = 5`.
The energy of the electron `= 100 eV` (given). The energy supplied to `He^(+)` ion = `52.3 eV`. Therefore , the energy of the electron left after the collision `= 100 - 52.3 = 47.7 eV`