In the spectrum of hydrogen atom, the ratio of the longest wavelength in Lyman series to the longest wavelangth in the Balmer series is

Options

A. `(5)/(27)`
B. `(1)/(93)`
C. `(4)/(9)`
D. `(3)/(2)`

Answer

To find the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series for the hydrogen atom, we can follow these steps:

### Step 1: Understand the Series
The Lyman series corresponds to transitions where the electron falls to the n=1 energy level, while the Balmer series corresponds to transitions where the electron falls to the n=2 energy level.

### Step 2: Identify the Longest Wavelength
The longest wavelength in a series corresponds to the lowest energy transition. For the Lyman series, the longest wavelength occurs when the electron transitions from n=2 to n=1. For the Balmer series, the longest wavelength occurs when the electron transitions from n=3 to n=2.

### Step 3: Use the Rydberg Formula
The Rydberg formula for the wavelength of the emitted light 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, \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.

### Step 4: Calculate the Wavelength for the Lyman Series
For the Lyman series (transition from n=2 to n=1):
- \( n_1 = 1 \)
- \( n_2 = 2 \)

Using the Rydberg formula:

\[ \frac{1}{\lambda_L} = R_H \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R_H \left( 1 - \frac{1}{4} \right) = R_H \left( \frac{3}{4} \right) \]

Thus,

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

### Step 5: Calculate the Wavelength for the Balmer Series
For the Balmer series (transition from n=3 to n=2):
- \( n_1 = 2 \)
- \( n_2 = 3 \)

Using the Rydberg formula:

\[ \frac{1}{\lambda_B} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R_H \left( \frac{1}{4} - \frac{1}{9} \right) = R_H \left( \frac{9 - 4}{36} \right) = R_H \left( \frac{5}{36} \right) \]

Thus,

\[ \lambda_B = \frac{36}{5 R_H} \]

### Step 6: Calculate the Ratio of Wavelengths
Now we can find the ratio of the longest wavelength in the Lyman series to that in the Balmer series:

\[ \text{Ratio} = \frac{\lambda_L}{\lambda_B} = \frac{\frac{4}{3 R_H}}{\frac{36}{5 R_H}} = \frac{4}{3} \cdot \frac{5}{36} = \frac{20}{108} = \frac{5}{27} \]

### Conclusion
The ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is:

\[ \frac{5}{27} \]