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

`DeltaE=13.6 Z^(2) [(1)/(1^(2))-(1)/((2n)^(2))]`
`204=13.6 Z^(2) (1-(1)/(4n^(2)))`
`DeltaE' =13.6 Z^(2) ((1)/(n^(2))-(1)/((2n)^(2)))`
`40.8=13.6 Z^(2) ((1)/(n^(2))-(1)/(4n^(2)))`
`(i)//(ii)`
`(204)/(40.8)=(1-1//4n^(2))/(1//n^(2)-1//4n^(2))`
`5((1)/(n^(2))-(1)/(4n^(2)))=1-(1)/(4n^(2))`
` (5)/(n^(2))-(4)/(4n^(2)) =1 implies (20-4)/(4n^(2))=1`
n=2
`204=13.6 Z^(2)(1-(1)/(4n^(2)))`
`=13.6 Z ^(2) (1-(1)/(4(2)^(2))=1-(1)/(16)=(15)/(16))`
`Z^(2)=(204xx16)/(13.6xx15)=16 implies Z=4`
`E_(n)=-13.6(Z^(2))/(n^(2)),` for ground state, n=1
` E_(1) =-13.6 (4)^(2)=-217.6 eV`
Minimum energy will be in the transition
`2n "to" (2n-1) i.e. 4 " to " 3 `
`Delta E=13.6 Z^(2) ((1)/(3^(2))-(1)/(4^(2)))`
`=13.6(4)^(2) ((1)/(9)-(1)/(16)=(7)/(144))`
`(13.6xx16xx7)/(144)`
10.6 eV