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.

Options

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

Answer

B

Explanation

Let ground state enegry (in `eV`) be `E_(1)` Then from the given condition
`E_(2n) - E_(1)= 204 ev` or `(E_(1))/(4n^(2)) - E_(1) = 204 eV`
`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))) = 40.8 eV .....(ii)`
From equations `(i)` and `(i)`, `(1-(1)/(4n^(2)))/((3)/(4n^(2))) = 5rArrn = 2`