In a transition to a state of ionization potential `30.6 V, (a)` hydrogen like atom emits a `541.17 A^o` photon. Determine the principal quantum number of the initial state.

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Given Data
We are given:
- Ionization potential (energy) \( E = 30.6 \, \text{eV} \)
- Wavelength of the emitted photon \( \lambda = 541.17 \, \text{Å} \)

### Step 2: Convert Wavelength to Meters
First, we need to convert the wavelength from Angstroms to meters:
\[ \lambda = 541.17 \, \text{Å} = 541.17 \times 10^{-10} \, \text{m} \]

### Step 3: Calculate the Rydberg Constant for Hydrogen-like Atoms
The ionization energy for a hydrogen-like atom is given by:
\[ E = 13.6 \, Z^2 \, \text{eV} \]
Setting this equal to the given ionization potential:
\[ 30.6 = 13.6 \, Z^2 \]
Solving for \( Z^2 \):
\[ Z^2 = \frac{30.6}{13.6} \approx 2.25 \]
Taking the square root gives:
\[ Z \approx 1.5 \]
Since \( Z \) must be a whole number, we round it to the nearest whole number, which is \( Z = 2 \).

### Step 4: Use the Rydberg Formula
The Rydberg formula for the wavelength of emitted photons in hydrogen-like atoms is:
\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
Where \( R \) is the Rydberg constant, approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \).

### Step 5: Substitute Known Values
Using \( Z = 2 \) and substituting the values into the Rydberg formula:
\[ \frac{1}{541.17 \times 10^{-10}} = (1.097 \times 10^7) \cdot (2^2) \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
This simplifies to:
\[ \frac{1}{541.17 \times 10^{-10}} = (1.097 \times 10^7) \cdot 4 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

### Step 6: Calculate the Left Side
Calculating the left-hand side:
\[ \frac{1}{541.17 \times 10^{-10}} \approx 1.847 \times 10^9 \, \text{m}^{-1} \]

### Step 7: Set Up the Equation
Setting the left side equal to the right side:
\[ 1.847 \times 10^9 = (1.097 \times 10^7) \cdot 4 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
Calculating the right side:
\[ 4 \cdot 1.097 \times 10^7 = 4.388 \times 10^7 \]
Thus:
\[ 1.847 \times 10^9 = 4.388 \times 10^7 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

### Step 8: Solve for \( n_1 \) and \( n_2 \)
Rearranging gives:
\[ \frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{1.847 \times 10^9}{4.388 \times 10^7} \]
Calculating the right side:
\[ \frac{1.847 \times 10^9}{4.388 \times 10^7} \approx 42.05 \]
So:
\[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \approx 42.05 \]

### Step 9: Assume \( n_2 = 1 \) (Ground State)
Assuming the final state \( n_2 = 1 \) (ground state):
\[ \frac{1}{n_1^2} - 1 = 42.05 \]
This implies:
\[ \frac{1}{n_1^2} = 43.05 \]
Thus:
\[ n_1^2 = \frac{1}{43.05} \approx 0.0232 \]
Taking the square root gives:
\[ n_1 \approx 6.56 \]
Rounding gives \( n_1 = 7 \).

### Final Answer
The principal quantum number of the initial state is \( n_1 = 7 \).

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