A hydrogen-like atom (aatomic number `Z`) is in a hight excited state of quantum number `n` . This excited atom can make a transition to the first excited state by successively emitting two proton of energy `10.20 e V and 17.00 e V`, respectively.
Alternatively , the atom from the same excited state can make a transition to the second excited state by successively emitting two proton of energy `4.25 e V and 5.954 e V`, respectively. Determine the value of `n and Z` (ionization energy of hydrogen 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 transitions and emitted photon energies provided.

### Step-by-Step Solution:

1. **Understanding the Transitions**:
- The atom transitions from a high excited state \( n \) to the first excited state \( n = 2 \) by emitting two photons with energies \( 10.20 \, \text{eV} \) and \( 17.00 \, \text{eV} \).
- The total energy emitted in this transition is:
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