Consider a hydrogen-like atom whose energy in nth excited state is given by
`E_(n) = -(13.6 Z^(2))/(n^(2))`
When this excited makes a transition from excited state to ground state , most energetic photons have energy
`E_(max) = 52.224 eV`. and least energetic photons have energy
`E_(max) = 1.224 eV`
Find the atomic number of atom and the intial state or excitation.

Explanation

To solve the problem, we need to find the atomic number \( Z \) and the initial state \( n_i \) of a hydrogen-like atom based on the given energies of the photons emitted during transitions from an excited state to the ground state.

### Step 1: Understand the Energy Levels
The energy of an electron in the \( n \)-th excited state of a hydrogen-like atom is given by the formula:
\[ E_n = -\frac{13.6 Z^2}{n^2} \text{ eV} \]
where \( Z \) is the atomic number and \( n \) is the principal quantum number.
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