The ratio of the longest to shortest wavelength in Brackett series of hydrogen spectra is

Options

A. `(25)/(9)`
B. `(17)/(6)`
C. `(9)/(5)`
D. `(4)/(3)`

Explanation

To find the ratio of the longest to shortest wavelength in the Brackett series of the hydrogen spectrum, we can follow these steps:

### Step 1: Identify the Brackett Series
The Brackett series corresponds to electronic transitions where the electron falls to the n=4 energy level (n1 = 4) from higher energy levels (n2 = 5, 6, 7, ...).

### Step 2: Determine the Longest Wavelength
The longest wavelength corresponds to the smallest energy transition, which occurs when the electron transitions from n=5 to n=4.

Using the formula for wavelength:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
For the longest wavelength (n2 = 5):
\[ \frac{1}{\lambda_{max}} = R \left( \frac{1}{4^2} - \frac{1}{5^2} \right) \]
Calculating this gives:
\[ \frac{1}{\lambda_{max}} = R \left( \frac{1}{16} - \frac{1}{25} \right) \]
Finding a common denominator (400):
\[ \frac{1}{\lambda_{max}} = R \left( \frac{25 - 16}{400} \right) = R \left( \frac{9}{400} \right) \]
Thus,
\[ \lambda_{max} = \frac{400}{9R} \]

### Step 3: Determine the Shortest Wavelength
The shortest wavelength corresponds to the largest energy transition, which occurs when the electron transitions from n=∞ to n=4.

Using the same formula:
\[ \frac{1}{\lambda_{min}} = R \left( \frac{1}{4^2} - \frac{1}{\infty^2} \right) \]
Since \( \frac{1}{\infty^2} = 0 \):
\[ \frac{1}{\lambda_{min}} = R \left( \frac{1}{16} \right) \]
Thus,
\[ \lambda_{min} = 16R \]

### Step 4: Calculate the Ratio of Longest to Shortest Wavelength
Now we can find the ratio of the longest wavelength to the shortest wavelength:
\[ \frac{\lambda_{max}}{\lambda_{min}} = \frac{\frac{400}{9R}}{16R} = \frac{400}{9R} \cdot \frac{1}{16R} = \frac{400}{144R^2} = \frac{25}{9} \]

### Final Result
The ratio of the longest to shortest wavelength in the Brackett series of hydrogen spectra is:
\[ \frac{25}{9} \]