The ratio of longest wavelength and the shortest wavelength observed in the five spectral series of emission spectrum of hydrogen is

Options

A. `(4)/(3)`
B. `(525)/(376)`
C. `25`
D. `(900)/(11)`

Explanation

To find the ratio of the longest wavelength (\( \lambda_{\text{max}} \)) and the shortest wavelength (\( \lambda_{\text{min}} \)) observed in the five spectral series of the emission spectrum of hydrogen, we can use the Rydberg formula:

\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

where:
- \( R \) is the Rydberg constant,
- \( n_1 \) and \( n_2 \) are the principal quantum numbers of the electron transitions.

### Step 1: Identify \( n_1 \) and \( n_2 \) for the shortest wavelength
The shortest wavelength corresponds to the highest energy transition, which occurs when \( n_1 = 1 \) and \( n_2 = \infty \).

Using the Rydberg formula for the shortest wavelength:

\[ \frac{1}{\lambda_{\text{min}}} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R \left( 1 - 0 \right) = R \]

Thus,

\[ \lambda_{\text{min}} = \frac{1}{R} \]

### Step 2: Identify \( n_1 \) and \( n_2 \) for the longest wavelength
The longest wavelength corresponds to the lowest energy transition in the series, which occurs when \( n_1 = 5 \) and \( n_2 = 6 \).

Using the Rydberg formula for the longest wavelength:

\[ \frac{1}{\lambda_{\text{max}}} = R \left( \frac{1}{5^2} - \frac{1}{6^2} \right) \]

Calculating \( \frac{1}{5^2} - \frac{1}{6^2} \):

\[ \frac{1}{5^2} = \frac{1}{25}, \quad \frac{1}{6^2} = \frac{1}{36} \]

Finding a common denominator (which is 900):

\[ \frac{1}{25} = \frac{36}{900}, \quad \frac{1}{36} = \frac{25}{900} \]

Thus,

\[ \frac{1}{\lambda_{\text{max}}} = R \left( \frac{36}{900} - \frac{25}{900} \right) = R \left( \frac{11}{900} \right) \]

Therefore,

\[ \lambda_{\text{max}} = \frac{900}{11R} \]

### Step 3: Calculate the ratio of longest to shortest wavelength
Now, we can find the ratio \( \frac{\lambda_{\text{max}}}{\lambda_{\text{min}}} \):

\[ \frac{\lambda_{\text{max}}}{\lambda_{\text{min}}} = \frac{\frac{900}{11R}}{\frac{1}{R}} = \frac{900}{11} \]

### Final Result
The ratio of the longest wavelength to the shortest wavelength observed in the five spectral series of the emission spectrum of hydrogen is:

\[ \frac{\lambda_{\text{max}}}{\lambda_{\text{min}}} = \frac{900}{11} \]