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)`

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 Lyman and Balmer Series
- The Lyman series corresponds to transitions where the electron falls to the n=1 energy level.
- 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 corresponds to the transition with the smallest energy difference, which occurs when the electron moves from the nearest higher energy level to the specified level.

### Step 3: Calculate the Longest Wavelength in the Lyman Series
- For the Lyman series, the transition is from n=2 to n=1.
- Using the Rydberg formula:
\[ \frac{1}{\lambda_L} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( n_1 = 1 \) and \( n_2 = 2 \):
\[ \frac{1}{\lambda_L} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right) \]
Therefore,
\[ \lambda_L = \frac{4}{3R} \]

### Step 4: Calculate the Longest Wavelength in the Balmer Series
- For the Balmer series, the transition is from n=3 to n=2.
- Again using the Rydberg formula:
\[ \frac{1}{\lambda_B} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( n_1 = 2 \) and \( n_2 = 3 \):
\[ \frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) \]
Finding a common denominator (36):
\[ \frac{1}{\lambda_B} = R \left( \frac{9}{36} - \frac{4}{36} \right) = R \left( \frac{5}{36} \right) \]
Therefore,
\[ \lambda_B = \frac{36}{5R} \]

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

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