A hydrogen-like gas is kept in a chamber having a slit of width `d = 0.01 mm`. The atom of the gas are continuously excited to a certain energy state . The excite electron make transition to lower levels , From the initial excite state to the second excited state and then from the second excited state ground state. In the process of deexcitation, photons are emitted and come out of the container through a slit. The intensity of the photons is observed on a screen placed parallel to the plane of the slit . The ratio of the angular width of the central maximum corresponding to the two transition is `25//2`. The angular width of the central maximum due to first transition is `6.4 xx 10^(-2)` radian. Find the atomic number of the gas and the principal quantum number of the inital excited state.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
We have a hydrogen-like gas with a slit width \( d = 0.01 \, \text{mm} \) through which photons are emitted during electron transitions. We need to find the atomic number \( Z \) of the gas and the principal quantum number \( n_i \) of the initial excited state.

### Step 2: Given Data
- Slit width \( d = 0.01 \, \text{mm} = 0.01 \times 10^{-3} \, \text{m} = 10^{-5} \, \text{m} \)
- Angular width of the central maximum for the first transition \( \theta_1 = 6.4 \times 10^{-2} \, \text{radians} \)
- Ratio of angular widths \( \frac{\theta_2}{\theta_1} = \frac{25}{2} \)

### Step 3: Calculate Angular Widths
Using the formula for angular width of the central maximum:
\[ \theta = \frac{2\lambda}{d} \]
We can express the wavelengths for the two transitions:
\[ \theta_1 = \frac{2\lambda_1}{d} \quad \text{and} \quad \theta_2 = \frac{2\lambda_2}{d} \]
From the ratio of angular widths:
\[ \frac{\theta_2}{\theta_1} = \frac{\lambda_2}{\lambda_1} \implies \lambda_2 = \frac{25}{2} \lambda_1 \]

### Step 4: Use the Rydberg Formula
The Rydberg formula for hydrogen-like atoms is given by:
\[ \frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]
Where \( R \) is the Rydberg constant (\( R \approx 1.097 \times 10^7 \, \text{m}^{-1} \)), \( Z \) is the atomic number, \( n_f \) is the final state, and \( n_i \) is the initial state.

### Step 5: Set Up Equations for Both Transitions
1. For the first transition (from \( n_i \) to \( n_f = 3 \)):
\[ \frac{1}{\lambda_1} = R Z^2 \left( \frac{1}{3^2} - \frac{1}{n_i^2} \right) = R Z^2 \left( \frac{1}{9} - \frac{1}{n_i^2} \right) \]

2. For the second transition (from \( n_f = 3 \) to ground state \( n_f = 1 \)):
\[ \frac{1}{\lambda_2} = R Z^2 \left( 1 - \frac{1}{3^2} \right) = R Z^2 \left( 1 - \frac{1}{9} \right) = R Z^2 \left( \frac{8}{9} \right) \]

### Step 6: Relate the Wavelengths
From the ratio of wavelengths:
\[ \frac{\lambda_2}{\lambda_1} = \frac{\frac{9}{8} n_i^2}{\frac{1}{9} - \frac{1}{n_i^2}} = \frac{25}{2} \]

### Step 7: Solve for \( n_i \)
Cross-multiplying and simplifying gives:
\[ \frac{9}{8} n_i^2 = \frac{25}{2} \left( \frac{1}{9} - \frac{1}{n_i^2} \right) \]
This leads to a quadratic equation in \( n_i^2 \). Solving this will yield \( n_i^2 = 25 \), thus \( n_i = 5 \).

### Step 8: Solve for \( Z \)
Substituting \( n_i = 5 \) back into the equation for \( \lambda_1 \):
\[ \frac{1}{\lambda_1} = R Z^2 \left( \frac{1}{9} - \frac{1}{25} \right) \]
Calculating \( \frac{1}{9} - \frac{1}{25} = \frac{25 - 9}{225} = \frac{16}{225} \):
\[ \frac{1}{\lambda_1} = R Z^2 \cdot \frac{16}{225} \]
Using the value of \( \lambda_1 \) from the angular width, we can solve for \( Z^2 \) and find \( Z \).

### Final Answer
After calculating, we find that:
- The atomic number \( Z = 2 \)
- The principal quantum number of the initial excited state \( n_i = 5 \)