The average lifetime for the `n = 3` excited state of a hydrogen-like atom is `4.8 xx 10^(-8) s` and that for the `n = 2 state is 12.8 xx 10^(-8) s`. The ratio of average number of revolution made in the `n = 3` sate before any transition can take place from these state is.

Explanation

To solve the problem, we need to find the ratio of the average number of revolutions made in the \( n = 3 \) state before any transition occurs, compared to the \( n = 2 \) state. Here’s how we can approach this step by step:

### Step 1: Understand the relationship between frequency and quantum number
The frequency \( f \) of an electron in a hydrogen-like atom is inversely proportional to the cube of the principal quantum number \( n \):
\[ f \propto \frac{1}{n^3} \]
This means that:
\[ \frac{f_3}{f_2} = \frac{n_2^3}{n_3^3} \]
where \( n_3 = 3 \) and \( n_2 = 2 \).

### Step 2: Calculate the ratio of frequencies
Using the values of \( n \):
\[ \frac{f_3}{f_2} = \frac{2^3}{3^3} = \frac{8}{27} \]

### Step 3: Calculate the average number of revolutions
The average number of revolutions \( N \) made before a transition occurs is given by:
\[ N = f \times T \]
where \( T \) is the average lifetime of the state. Therefore, we can express the average number of revolutions for each state as:
\[ N_3 = f_3 \times T_3 \]
\[ N_2 = f_2 \times T_2 \]

### Step 4: Find the ratio of the average number of revolutions
Now, we can find the ratio \( \frac{N_3}{N_2} \):
\[ \frac{N_3}{N_2} = \frac{f_3 \times T_3}{f_2 \times T_2} \]
Substituting the expression for the frequencies:
\[ \frac{N_3}{N_2} = \frac{f_3}{f_2} \times \frac{T_3}{T_2} = \frac{8/27}{T_2/T_3} \]

### Step 5: Substitute the average lifetimes
Given:
- \( T_3 = 4.8 \times 10^{-8} \, s \)
- \( T_2 = 12.8 \times 10^{-8} \, s \)

Now substituting these values into the equation:
\[ \frac{N_3}{N_2} = \frac{8}{27} \times \frac{4.8 \times 10^{-8}}{12.8 \times 10^{-8}} \]
This simplifies to:
\[ \frac{N_3}{N_2} = \frac{8}{27} \times \frac{4.8}{12.8} \]

### Step 6: Calculate the numerical value
Calculating \( \frac{4.8}{12.8} \):
\[ \frac{4.8}{12.8} = \frac{48}{128} = \frac{3}{8} \]
Now substituting back:
\[ \frac{N_3}{N_2} = \frac{8}{27} \times \frac{3}{8} = \frac{3}{27} = \frac{1}{9} \]

### Final Result
Thus, the ratio of the average number of revolutions made in the \( n = 3 \) state before any transition can take place is:
\[ \frac{N_3}{N_2} = 9 \]