The ratio of wavelength values of series limit lines `(n_(2) = oo)` of Balmer series and Paschen series are

Options

A. `4:9`
B. `9:4`
C. `2:3`
D. `3:2`

Answer

To find the ratio of the wavelength values of the series limit lines (where \( n_2 = \infty \)) of the Balmer series and the Paschen series, we can use the Rydberg formula for hydrogen:

\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

where \( R_H \) is the Rydberg constant, \( n_1 \) is the principal quantum number of the lower energy level, and \( n_2 \) is the principal quantum number of the higher energy level.

### Step 1: Determine the values of \( n_1 \) and \( n_2 \) for both series.

- For the **Balmer series**, the lower level \( n_1 = 2 \) and the upper level \( n_2 = \infty \).
- For the **Paschen series**, the lower level \( n_1 = 3 \) and the upper level \( n_2 = \infty \).

### Step 2: Apply the Rydberg formula for the Balmer series.

Using the Rydberg formula for the Balmer series:

\[ \frac{1}{\lambda_B} = R_H \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) \]

Since \( \frac{1}{\infty^2} = 0 \), we have:

\[ \frac{1}{\lambda_B} = R_H \left( \frac{1}{4} \right) \]

Thus,

\[ \lambda_B = \frac{4}{R_H} \]

### Step 3: Apply the Rydberg formula for the Paschen series.

Now, using the Rydberg formula for the Paschen series:

\[ \frac{1}{\lambda_P} = R_H \left( \frac{1}{3^2} - \frac{1}{\infty^2} \right) \]

Again, since \( \frac{1}{\infty^2} = 0 \), we have:

\[ \frac{1}{\lambda_P} = R_H \left( \frac{1}{9} \right) \]

Thus,

\[ \lambda_P = \frac{9}{R_H} \]

### Step 4: Calculate the ratio of the wavelengths.

Now, we can find the ratio of the wavelengths \( \lambda_B \) and \( \lambda_P \):

\[ \frac{\lambda_B}{\lambda_P} = \frac{\frac{4}{R_H}}{\frac{9}{R_H}} = \frac{4}{9} \]

### Conclusion:

The ratio of the wavelength values of the series limit lines of the Balmer series to the Paschen series is:

\[ \frac{\lambda_B}{\lambda_P} = \frac{4}{9} \]

### Final Result:

The answer is \( \frac{4}{9} \).

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