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 eV) be `E_(1)`
Then from the given condition
`E_(2n)-E_(1)=204eV` or `(E_(1))/(4n^(2))-E_(1)=204eV`
`rArr E_(1)((1)/(4n^(2))-1)=204eV`...(i)
and `E_(2n) -E_(n) = 40.8 eV`
`rArr (E_(1))/(4n^(2))-(E_(1))/(n^(2))=E_(1)(-(3)/(4n^(2)))=40.8eV`....(ii)
From equation (i) and (ii),
`(1-(1)/(4n^(2)))/((3)/(4n^(2)))=5 rArr n=2` .