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.

Answer

4

Explanation

Let ground state energy (in eV) be `E_(1)`
Then, from the given condition
`E_(2n)-E_(1)=204 eV`
or `(E_(1))/(4n^(2))-E_(1)=204eV`