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 emitting a photon of first `27.2eV`. Alternatively the atom from the same excited state can make a transition of energy `10.20eV` the value of `n` and `z` are given (ionization energy of hydrogen atom is 13.6eV)

Options

A. `n==6` and `z=3`
B. `n=3` and `z=6`
C. `n=8` and `z=4`
D. `n=4` and `z=8`

Explanation

To solve the problem, we will use the energy level formula for hydrogen-like atoms and the given transition energies to find the values of \( n \) and \( Z \).

### Step-by-Step Solution:

1. **Understanding the Energy Levels**:
The energy of the nth level of a hydrogen-like atom is given by:
\[ E_n = -\frac{13.6 Z^2}{n^2} \text{ eV} \]
where \( Z \) is the atomic number.

2. **Transition Energy**:
The energy difference between two levels \( n_1 \) and \( n_2 \) is given by:
\[ \Delta E = E_{n_1} - E_{n_2} = 13.6 Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

3. **First Transition**:
The atom is in a higher excited state \( n \) and transitions to the first excited state \( n_1 = 2 \) with an energy emission of \( 27.2 \text{ eV} \):
\[ 27.2 = 13.6 Z^2 \left( \frac{1}{2^2} - \frac{1}{n^2} \right) \]
Simplifying this:
\[ 27.2 = 13.6 Z^2 \left( \frac{1}{4} - \frac{1}{n^2} \right) \]
Dividing both sides by \( 13.6 \):
\[ 2 = Z^2 \left( \frac{1}{4} - \frac{1}{n^2} \right) \]
Rearranging gives:
\[ \frac{1}{4} - \frac{1}{n^2} = \frac{2}{Z^2} \]

4. **Second Transition**:
The atom transitions to the second excited state \( n_1 = 3 \) with an energy emission of \( 10.20 \text{ eV} \):
\[ 10.20 = 13.6 Z^2 \left( \frac{1}{3^2} - \frac{1}{n^2} \right) \]
Simplifying this:
\[ 10.20 = 13.6 Z^2 \left( \frac{1}{9} - \frac{1}{n^2} \right) \]
Dividing both sides by \( 13.6 \):
\[ 0.75 = Z^2 \left( \frac{1}{9} - \frac{1}{n^2} \right) \]
Rearranging gives:
\[ \frac{1}{9} - \frac{1}{n^2} = \frac{0.75}{Z^2} \]

5. **Setting Up the Equations**:
We now have two equations:
- From the first transition:
\[ \frac{1}{4} - \frac{1}{n^2} = \frac{2}{Z^2} \quad \text{(1)} \]
- From the second transition:
\[ \frac{1}{9} - \frac{1}{n^2} = \frac{0.75}{Z^2} \quad \text{(2)} \]

6. **Eliminating \( \frac{1}{n^2} \)**:
Subtract equation (1) from equation (2):
\[ \left( \frac{1}{9} - \frac{1}{4} \right) = \left( \frac{0.75}{Z^2} - \frac{2}{Z^2} \right) \]
Simplifying the left side:
\[ \frac{1}{9} - \frac{1}{4} = \frac{4 - 9}{36} = -\frac{5}{36} \]
Simplifying the right side:
\[ \frac{0.75 - 2}{Z^2} = -\frac{1.25}{Z^2} \]
Setting them equal gives:
\[ -\frac{5}{36} = -\frac{1.25}{Z^2} \]
Cross-multiplying:
\[ 5Z^2 = 1.25 \times 36 \]
\[ 5Z^2 = 45 \Rightarrow Z^2 = 9 \Rightarrow Z = 3 \]

7. **Finding \( n \)**:
Substitute \( Z = 3 \) back into either equation to find \( n \). Using equation (1):
\[ \frac{1}{4} - \frac{1}{n^2} = \frac{2}{3^2} = \frac{2}{9} \]
Rearranging gives:
\[ \frac{1}{n^2} = \frac{1}{4} - \frac{2}{9} \]
Finding a common denominator (36):
\[ \frac{1}{n^2} = \frac{9}{36} - \frac{8}{36} = \frac{1}{36} \]
Thus:
\[ n^2 = 36 \Rightarrow n = 6 \]

### Final Answer:
The values are \( n = 6 \) and \( Z = 3 \).