Question

A hydrogen like atom with atomic number Z is in an excited state of quantum number 2n. It can emit a maximum energy photon of 204 eV. If it makes a transition ot quantum state n, a photon of energy 40.8 eV is emitted. Find n, Z and the ground state energy (in eV) of this atom. Also calculate the minimum energy (in eV) that can be emitted by this atom during de-excitation. Ground state energy of hydrogen atom is – 13. 6 eV.

Answer 1

The energy released during de-excitation n hydrogen like atoms is given by : `E_(n_(2)) - E_(n_(1)) = (me^(4))/(8epsi_(0)^(2)h^(2))[1/(n_(1)^(2))-(1)/(n_(2)^(2))]Z^(2)`
Energy released in de-excitation will be meximum if trannasation taken place from nth energy levwel to ground state
i.e.
`E_(2n) - E_(1) = 13.6[1/1^(2) - (1)/((2n)^(2))]Z^(2) = 204 eV`....(i) & also `E_(2n) - E_(n) = 13.5[1/n^(2) - 1/((2n)^(2)) ] 2^(2) = 40.8 eV` .....(ii)
Taking ratio of (i) to (ii), we w3ill get `(4n^(2) - 1)/(5) = 5 rArr n^(2) = 4 rArr n = 2`
Putting `n = 2` in equation (i) we ge `2^(2) = (204 xx 16)/(13.6 xx 15) rArr Z = 4`
`:. E_(n) = -13.6 (Z^(2))/(n^(2)) rArr E_(1) = - 13.6 xx (4^(2))/(1^(2)) = - 217.6 eV = "ground state energy"`
`Delta E` is minimum if transation will be from `2n` to `2n -1` i.e. between last two adjecvent energy levels.
`:. Delta E_(min) = E_(2n) - E_(2n - 1) = 13.6 [1/(3^(2)) - 1/(4^(2))] 4^(2) = 10.57 eV`
Is the minimum amount of energy released during do-excation.