A hydrogen like atom (atomic number `z`) is in a higher excited state of quantum number `n`. This excited atom can make a transition to the first excited state by successively emitting two photons of energies `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 2 photons of energy `4.25 eV` and`5.95 eV` respectively. Determine the value of `(n+z)`

Explanation

To solve the problem step by step, we need to analyze the transitions of the hydrogen-like atom and use the energy level formula for hydrogen-like atoms to find the values of \( n \) and \( z \).

### Step 1: Understand the Energy Transition Formula
The energy difference between two states in a hydrogen-like atom is given by the formula:
\[ \Delta E = -13.6 \, \text{eV} \cdot z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \( z \) is the atomic number,
- \( n_1 \) is the principal quantum number of the lower energy state,
- \( n_2 \) is the principal quantum number of the higher energy state.

### Step 2: First Transition to the First Excited State
The atom transitions from the excited state \( n \) to the first excited state \( n_1 = 2 \) by emitting two photons with energies \( 10.2 \, \text{eV} \) and \( 17.0 \, \text{eV} \). The total energy emitted is:
\[ E_1 = 10.2 \, \text{eV} + 17.0 \, \text{eV} = 27.2 \, \text{eV} \]
Using the energy transition formula:
\[ 27.2 = 13.6 z^2 \left( \frac{1}{2^2} - \frac{1}{n^2} \right) \]
This simplifies to:
\[ 27.2 = 13.6 z^2 \left( \frac{1}{4} - \frac{1}{n^2} \right) \]
Dividing both sides by \( 13.6 \):
\[ 2 = z^2 \left( \frac{1}{4} - \frac{1}{n^2} \right) \]
Let this be Equation (1).

### Step 3: Second Transition to the Second Excited State
The atom also transitions from the excited state \( n \) to the second excited state \( n_1 = 3 \) by emitting two photons with energies \( 4.25 \, \text{eV} \) and \( 5.95 \, \text{eV} \). The total energy emitted is:
\[ E_2 = 4.25 \, \text{eV} + 5.95 \, \text{eV} = 10.2 \, \text{eV} \]
Using the energy transition formula:
\[ 10.2 = 13.6 z^2 \left( \frac{1}{3^2} - \frac{1}{n^2} \right) \]
This simplifies to:
\[ 10.2 = 13.6 z^2 \left( \frac{1}{9} - \frac{1}{n^2} \right) \]
Dividing both sides by \( 13.6 \):
\[ 0.75 = z^2 \left( \frac{1}{9} - \frac{1}{n^2} \right) \]
Let this be Equation (2).

### Step 4: Solve the System of Equations
Now we have two equations:
1. \( 2 = z^2 \left( \frac{1}{4} - \frac{1}{n^2} \right) \)
2. \( 0.75 = z^2 \left( \frac{1}{9} - \frac{1}{n^2} \right) \)

From Equation (1):
\[ \frac{1}{n^2} = \frac{1}{4} - \frac{2}{z^2} \]
From Equation (2):
\[ \frac{1}{n^2} = \frac{1}{9} - \frac{0.75}{z^2} \]

Setting both expressions for \( \frac{1}{n^2} \) equal to each other:
\[ \frac{1}{4} - \frac{2}{z^2} = \frac{1}{9} - \frac{0.75}{z^2} \]

### Step 5: Solve for \( z^2 \)
Rearranging gives:
\[ \frac{2}{z^2} - \frac{0.75}{z^2} = \frac{1}{4} - \frac{1}{9} \]
Combine the left side:
\[ \frac{1.25}{z^2} = \frac{9 - 4}{36} = \frac{5}{36} \]
Cross-multiplying gives:
\[ 1.25 \cdot 36 = 5 z^2 \]
\[ 45 = 5 z^2 \implies z^2 = 9 \implies z = 3 \]

### Step 6: Substitute \( z \) Back to Find \( n \)
Substituting \( z = 3 \) back into Equation (1):
\[ 2 = 9 \left( \frac{1}{4} - \frac{1}{n^2} \right) \]
\[ 2 = \frac{9}{4} - \frac{9}{n^2} \]
Multiplying through by \( n^2 \):
\[ 2n^2 = \frac{9n^2}{4} - 9 \]
Rearranging gives:
\[ 8n^2 = 9n^2 - 36 \implies n^2 = 36 \implies n = 6 \]

### Final Calculation of \( n + z \)
Now we have \( n = 6 \) and \( z = 3 \):
\[ n + z = 6 + 3 = 9 \]

### Final Answer
The value of \( n + z \) is \( \boxed{9} \).