The ratio of wavelength of 1st line of Balmer series and 2nd line of Lyman series is :

Options

A. 32/5
B. 2
C. 3
D. 16

Answer

To find the ratio of the wavelength of the first line of the Balmer series to the second line of the Lyman series, we will follow these steps:

### Step 1: Understand the Series
- The **Balmer series** corresponds to transitions where electrons fall to the n=2 energy level from higher levels (n=3, 4, 5, ...).
- The **Lyman series** corresponds to transitions where electrons fall to the n=1 energy level from higher levels (n=2, 3, 4, ...).

### Step 2: Identify the Lines
- The **first line of the Balmer series** corresponds to the transition from n=3 to n=2.
- The **second line of the Lyman series** corresponds to the transition from n=3 to n=1.

### Step 3: Use the Rydberg Formula
The Rydberg formula for the wavelength (λ) of the emitted light during these transitions is given by:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R_H \) is the Rydberg constant.

### Step 4: Calculate the Wavelength for Balmer Series
For the first line of the Balmer series (n=3 to n=2):
\[ \frac{1}{\lambda_B} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R_H \left( \frac{1}{4} - \frac{1}{9} \right) \]
Calculating the fractions:
\[ \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} \]
Thus,
\[ \frac{1}{\lambda_B} = R_H \cdot \frac{5}{36} \quad \Rightarrow \quad \lambda_B = \frac{36}{5R_H} \]

### Step 5: Calculate the Wavelength for Lyman Series
For the second line of the Lyman series (n=3 to n=1):
\[ \frac{1}{\lambda_L} = R_H \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = R_H \left( 1 - \frac{1}{9} \right) \]
Calculating the fractions:
\[ 1 - \frac{1}{9} = \frac{9 - 1}{9} = \frac{8}{9} \]
Thus,
\[ \frac{1}{\lambda_L} = R_H \cdot \frac{8}{9} \quad \Rightarrow \quad \lambda_L = \frac{9}{8R_H} \]

### Step 6: Calculate the Ratio of Wavelengths
Now we need to find the ratio of the wavelengths:
\[ \frac{\lambda_B}{\lambda_L} = \frac{\frac{36}{5R_H}}{\frac{9}{8R_H}} = \frac{36}{5} \cdot \frac{8}{9} = \frac{36 \cdot 8}{5 \cdot 9} \]
Calculating this gives:
\[ \frac{288}{45} = \frac{32}{5} \]

### Final Answer
Thus, the ratio of the wavelength of the first line of the Balmer series to the second line of the Lyman series is:
\[ \frac{\lambda_B}{\lambda_L} = \frac{32}{5} \]