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.

Explanation

`E_(2n)-E_(1)=204 eV`
Using the relation ,
`DeltaE=-13.6Z^(2)((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))`
We get, `(13.6)Z^(2) (13.63)Z^(2)(1-(1)/(4n^(2)))=204" "....(i)`
`E_(2n)-E_(n)=40.8 eV`
`:. 13.6 Z^(2)((1)/(n^(2))-(1)/(4n^(2)))=40.8" " ....(ii)`
Soving Eqs. (i) and (ii) we get, n=2 and Z=4
We know that, in gound state `(n=1),E=-13.6 Z^(2) eV`
SO, `" " E_(1)=(-13.6)Z^(2) eV`
`=-(13.6)(4)^(2) eV`
During de-excitation, minimum energy emitted is ,
`E_("min")=E_(2n)-E_(2n-1)=E_(4)-E_(3)`
`=(-217.6) /(4^(2))=((-217.6)/(3^(3)))=10.58 eV`