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.

Explanation

To solve the problem step by step, we will analyze the given information and apply the relevant formulas.

### Step 1: Understanding the Given Information
We have a hydrogen-like atom with atomic number \( Z \) in an excited state with quantum number \( 2n \). The maximum energy photon emitted is \( 204 \, \text{eV} \), and when transitioning to quantum state \( n \), a photon of energy \( 40.8 \, \text{eV} \) is emitted.

### Step 2: Energy Levels of Hydrogen-like Atoms
The energy of an electron in a hydrogen-like atom is given by the formula:
\[
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