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. `(3)/(23)`
C. `(7)/(29)`
D. `(9)/(31)`

Explanation

To solve the problem of finding 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 final energy level is n=1, while the Balmer series corresponds to transitions where the final energy level is n=2. The longest wavelength in each series corresponds to the transition with the smallest energy difference, which occurs between the closest energy levels.

### Step 2: Identify the Transitions
- For the **Lyman series**, the longest wavelength corresponds to the transition from n=2 to n=1.
- For the **Balmer series**, the longest wavelength corresponds to the transition from n=3 to n=2.

### Step 3: Use the Rydberg Formula
The wavelength (λ) of the emitted light can be calculated using the Rydberg formula:
\[ \frac{1}{\lambda} = R \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
Where:
- \( R \) is the Rydberg constant,
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)),
- \( n_1 \) and \( n_2 \) are the principal quantum numbers of the lower and upper energy levels respectively.

### Step 4: Calculate the Longest Wavelength for Lyman Series
For the Lyman series (n=2 to n=1):
\[ \frac{1}{\lambda_{L}} = R \cdot 1^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \cdot \frac{3}{4} \]
Thus,
\[ \lambda_{L} = \frac{4}{3R} \]

### Step 5: Calculate the Longest Wavelength for Balmer Series
For the Balmer series (n=3 to n=2):
\[ \frac{1}{\lambda_{B}} = R \cdot 1^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) \]
Calculating the difference:
\[ \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} \]
Thus,
\[ \lambda_{B} = \frac{36}{5R} \]

### Step 6: Calculate the Ratio of Wavelengths
Now we need to find the ratio of the longest wavelength in the Lyman series to that in the Balmer series:
\[ \frac{\lambda_{L}}{\lambda_{B}} = \frac{\frac{4}{3R}}{\frac{36}{5R}} = \frac{4}{3} \cdot \frac{5}{36} = \frac{20}{108} = \frac{5}{27} \]

### Final Answer
Thus, the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is:
\[ \frac{5}{27} \]