A hydrogen like atom (atomic number Z) is in a higher excited state of quantum number n. The excited atom can make a transition ot the first excited state by successively emitting two photons of energy 10.2 eV and 17.0 eV, respectively. Alternatively, the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energies 4.25 eV and 5.95 eV, respectivley Determine the values of n and Z. (lonization energy of H-atom = 13.6 eV)

Explanation

To solve the problem, we need to determine the quantum number \( n \) and the atomic number \( Z \) of a hydrogen-like atom based on the energy transitions provided.

### Step 1: Understand the Energy Transitions
The atom can transition from a higher excited state \( n \) to the first excited state (which corresponds to \( n=2 \)) by emitting two photons with energies \( 10.2 \, \text{eV} \) and \( 17.0 \, \text{eV} \).

The total energy emitted during this transition can be calculated as:
\[
E_1 = 10.2 \, \text{eV} + 17.0 \, \text{eV} = 27.2 \, \text{eV}
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