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 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 (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 energy (in `eV)` be `E_(1)`
Then from the given condition
`E_(2n)-E_(1) = 204eV` or `(E_(1))/(4n^(2)) - E_(1) = 204 eV`
`rArr E_(1) ((1)/(4n^(2))-1) = 204 eV` …….(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 (ii), `(1-(1)/(4n^(2)))`