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

Options

A. 1
B. 2
C. 3
D. 4

Answer

B

Explanation

Let ground state energy (in V) be `E_(1)`
Then from the given condition,
`" " E_(2n)-E_(1)=204eVor(E_(1))/(4n^(2))-E_(1)=204eV`
`rArr" " E_(1)((1)/(4n^(2))-1)=204eV" " ....(i)`
`and " " E_(2n)-E_(n)=40.eV`
`rArr(E_(1))/(4n^(2))-(E_(1))/(n^(2))=E_(1)(1-(3)/(4n^(2)))=40.8eV" ".....(ii)`
For Eqs. (i) and (ii) we get
`" " (1-(1)/(4n^(2)))/(3//4n^(2))=5rArrn=2`