The ratio of wavelength of a photon one present in limiting line of Balmer series and another with the longest wavelength line in the Lyman series in H atom is

Options

A. `3 : 1`
B. `2 : 1`
C. `4 : 3`
D. `2 : 3`

Answer

To solve the problem of finding the ratio of the wavelength of the limiting line of the Balmer series to the longest wavelength line in the Lyman series for a hydrogen atom, we can follow these steps:

### Step 1: Understand the Limiting Line of the Balmer Series
The limiting line of the Balmer series corresponds to the transition where the electron moves from an infinite energy level (n2 = ∞) to n1 = 2. The formula for the wavelength (λ) in terms of the Rydberg constant (R) is given by:

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

For the limiting line of the Balmer series:
- \( n_1 = 2 \)
- \( n_2 = \infty \)

Substituting these values into the formula:

\[ \frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - 0 \right) = R \left( \frac{1}{4} \right) \]

Thus, we have:

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

### Step 2: Understand the Longest Wavelength Line of the Lyman Series
The longest wavelength line in the Lyman series corresponds to the transition from n2 = 2 to n1 = 1. Again, using the Rydberg formula:

For the longest wavelength in the Lyman series:
- \( n_1 = 1 \)
- \( n_2 = 2 \)

Substituting these values into the 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, we have:

\[ \lambda_L = \frac{4}{3R} \]

### Step 3: Calculate the Ratio of the Wavelengths
Now we can find the ratio of the wavelengths of the limiting line of the Balmer series to the longest wavelength line in the Lyman series:

\[ \text{Ratio} = \frac{\lambda_B}{\lambda_L} = \frac{\frac{4}{R}}{\frac{4}{3R}} = \frac{4}{R} \times \frac{3R}{4} = 3 \]

Thus, the ratio of the wavelengths is:

\[ \text{Ratio} = 3:1 \]

### Final Answer
The ratio of the wavelength of the photon in the limiting line of the Balmer series to the longest wavelength line in the Lyman series in a hydrogen atom is **3:1**.

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