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. `(4)/(9)`
B. `(9)/(4)`
C. `(27)/(5)`
D. `(5)/(27)`

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-by-Step Solution:

1. **Understand the Series**:
- The Lyman series corresponds to transitions where the final energy level (n_final) is 1.
- The Balmer series corresponds to transitions where the final energy level (n_final) is 2.

2. **Use the Wavelength Formula**:
The formula for the wavelength (λ) of the emitted or absorbed light during these transitions is given by:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_{\text{final}}^2} - \frac{1}{n_{\text{initial}}^2} \right) \]
where \( R \) is the Rydberg constant.

3. **Calculate the Longest Wavelength in the Lyman Series**:
- For the longest wavelength in the Lyman series, the transition occurs from n=2 to n=1.
- Substitute \( n_{\text{final}} = 1 \) and \( n_{\text{initial}} = 2 \):
\[ \frac{1}{\lambda_1} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right) \]
- Therefore,
\[ \lambda_1 = \frac{4}{3R} \]

4. **Calculate the Longest Wavelength in the Balmer Series**:
- For the longest wavelength in the Balmer series, the transition occurs from n=3 to n=2.
- Substitute \( n_{\text{final}} = 2 \) and \( n_{\text{initial}} = 3 \):
\[ \frac{1}{\lambda_2} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) \]
- To combine these fractions, find a common denominator:
\[ \frac{1}{\lambda_2} = R \left( \frac{9 - 4}{36} \right) = R \left( \frac{5}{36} \right) \]
- Therefore,
\[ \lambda_2 = \frac{36}{5R} \]

5. **Calculate the Ratio of the Wavelengths**:
- Now, we need to find the ratio \( \frac{\lambda_1}{\lambda_2} \):
\[ \frac{\lambda_1}{\lambda_2} = \frac{\frac{4}{3R}}{\frac{36}{5R}} = \frac{4}{3R} \times \frac{5R}{36} = \frac{4 \times 5}{3 \times 36} = \frac{20}{108} = \frac{5}{27} \]

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