A hydrogen like atom of 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 to 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(eV) that can be emitted by this atom during de - excitation. Ground state energy of hydrogen atom is -13.6 eV

Explanation

The energy released during de-excitation in hydrogen like atoms is given by : `E_(n_2)-E_(n_1)=(me^4)/(8epsilon _0^2h^2)[1/n_1^2 -1/n_2^2]Z^2`
Energy released in de-excitation will be maximum if transition takes place from nth energy level to ground state.
`i.e., E_(2n)-E_1 =13.6 [1/1^2-1/(2n)^2]Z^2=204` eV....(i)
and also
`E_(2n)-E_n =13.6 [1/n^2-1/(2n)^2]Z^2=40.8` eV....(ii)
Taking ratio of (i) to (ii) , we will get
`(4n^2-1)/3=5 rArr n^2=4`
`rArr` n=2
Putting n=2 in equation (i) we get
`Z^2=(204xx16)/(13.6xx15)`
`rArr` Z=4
`because E_n=-13.6 Z^2/n^2`
`rArr E_1=-13.6xx4^2/1^2=-217.6` eV
= ground state energy
`DeltaE ` is minimum if transition will be form 2n to 2n-1. i.e., between last two adjacent energy levels.
`therefore DeltaE_"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 de-excitation.