Ratio of the wavelengths of first line of Lyman series and first line of Balmer series is 5 : X then find X.

Explanation

To find the value of \( X \) in the ratio of the wavelengths of the first line of the Lyman series and the first line of the Balmer series, we will follow these steps:

### Step 1: Understand the transitions for the series
- The first line of the Lyman series corresponds to the transition from \( n = 2 \) to \( n = 1 \).
- The first line of the Balmer series corresponds to the transition from \( n = 3 \) to \( n = 2 \).

### Step 2: Use the Rydberg formula
The Rydberg formula for the wavelength \( \lambda \) is given by:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]
where \( R \) is the Rydberg constant, \( n_f \) is the final energy level, and \( n_i \) is the initial energy level.

### Step 3: Calculate the wavelength for the Lyman series
For the Lyman series (transition from \( n = 2 \) to \( n = 1 \)):
- \( n_f = 1 \)
- \( n_i = 2 \)

Using the Rydberg formula:
\[ \frac{1}{\lambda_L} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right) \]
Thus,
\[ \lambda_L = \frac{4}{3R} \]

### Step 4: Calculate the wavelength for the Balmer series
For the Balmer series (transition from \( n = 3 \) to \( n = 2 \)):
- \( n_f = 2 \)
- \( n_i = 3 \)

Using the Rydberg formula:
\[ \frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) \]
Finding a common denominator (36):
\[ \frac{1}{\lambda_B} = R \left( \frac{9}{36} - \frac{4}{36} \right) = R \left( \frac{5}{36} \right) \]
Thus,
\[ \lambda_B = \frac{36}{5R} \]

### Step 5: Find the ratio of the wavelengths
Now, we need to find the ratio of the wavelengths:
\[ \frac{\lambda_L}{\lambda_B} = \frac{\frac{4}{3R}}{\frac{36}{5R}} = \frac{4}{3} \cdot \frac{5}{36} = \frac{20}{108} = \frac{5}{27} \]

### Step 6: Set up the ratio
According to the problem, the ratio of the wavelengths is given as:
\[ \frac{\lambda_L}{\lambda_B} = \frac{5}{X} \]
From our calculation, we have:
\[ \frac{5}{27} = \frac{5}{X} \]

### Step 7: Solve for \( X \)
By equating the two ratios:
\[ X = 27 \]

Thus, the final answer is:
\[ \boxed{27} \]