In the spectrum of hydrogen atom, the ratio of the longest wavelength in Lyman series to the longest wavelength 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 a hydrogen atom, we can follow these steps:

### Step 1: Identify the transitions for the longest wavelengths
- The longest wavelength in the Lyman series corresponds to the transition from \( n = 2 \) to \( n = 1 \).
- The longest wavelength in the Balmer series corresponds to the transition from \( n = 3 \) to \( n = 2 \).

### Step 2: Use the Rydberg formula for wavelengths
The Rydberg formula for the wavelength of emitted light during electron transitions in a hydrogen atom is given by:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R \) is the Rydberg constant, \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.

### Step 3: Calculate the longest wavelength in the Lyman series
For the Lyman series transition from \( n = 2 \) to \( n = 1 \):
\[ \frac{1}{\lambda_L} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right) \]
Thus, the wavelength \( \lambda_L \) is:
\[ \lambda_L = \frac{4}{3R} \]

### Step 4: Calculate the longest wavelength in the Balmer series
For the Balmer series transition from \( n = 3 \) to \( n = 2 \):
\[ \frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{9 - 4}{36} \right) = R \left( \frac{5}{36} \right) \]
Thus, the wavelength \( \lambda_B \) is:
\[ \lambda_B = \frac{36}{5R} \]

### Step 5: Calculate the ratio of the wavelengths
Now, we can find the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series:
\[ \frac{\lambda_L}{\lambda_B} = \frac{\frac{4}{3R}}{\frac{36}{5R}} = \frac{4}{3} \cdot \frac{5}{36} = \frac{20}{108} = \frac{5}{27} \]

### Conclusion
The ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is:
\[ \frac{\lambda_L}{\lambda_B} = \frac{5}{27} \]