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`

Explanation

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 first energy level (n=1) from higher levels (n=2, 3, 4, ...).
- The Balmer series corresponds to transitions where the electron falls to the second energy level (n=2) from higher levels (n=3, 4, 5, ...).

### Step 2: Identify the Longest Wavelength
- The longest wavelength corresponds to the smallest 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 (λ) of the emitted light 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.

### Step 4: Calculate the Longest Wavelength for Lyman Series
- For Lyman series (n1=1, n2=2):
\[ \frac{1}{\lambda_{L}} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right) \]
Thus,
\[ \lambda_{L} = \frac{4}{3R} \]

### Step 5: Calculate the Longest Wavelength for Balmer Series
- For Balmer series (n1=2, n2=3):
\[ \frac{1}{\lambda_{B}} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{9 - 4}{36} \right) = R \left( \frac{5}{36} \right) \]
Thus,
\[ \lambda_{B} = \frac{36}{5R} \]

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

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

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