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 solve the problem of finding the ratio of the longest to shortest wavelength in the Brackett series of hydrogen spectra, we can follow these steps:

### Step 1: Identify the series and the quantum numbers
In the Brackett series of hydrogen, the transitions occur from higher energy levels (n2) to n1 = 4. The longest wavelength corresponds to the transition from n2 = 5 to n1 = 4, and the shortest wavelength corresponds to the transition from n2 = ∞ to n1 = 4.

### Step 2: Calculate the longest wavelength (λ_max)
Using the Rydberg formula for the wavelength:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
For the longest wavelength:
- \( n_1 = 4 \)
- \( n_2 = 5 \)

Substituting these values into the formula:
\[ \frac{1}{\lambda_{max}} = R \left( \frac{1}{4^2} - \frac{1}{5^2} \right) \]
Calculating the squares:
\[ \frac{1}{\lambda_{max}} = R \left( \frac{1}{16} - \frac{1}{25} \right) \]
Finding a common denominator (LCM of 16 and 25 is 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: Calculate the shortest wavelength (λ_min)
For the shortest wavelength:
- \( n_1 = 4 \)
- \( n_2 = \infty \)

Using the Rydberg formula:
\[ \frac{1}{\lambda_{min}} = R \left( \frac{1}{4^2} - \frac{1}{\infty^2} \right) \]
Since \( \frac{1}{\infty^2} = 0 \), we have:
\[ \frac{1}{\lambda_{min}} = R \left( \frac{1}{16} \right) \]
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:
\[ \frac{\lambda_{max}}{\lambda_{min}} = \frac{\frac{400}{9R}}{16R} \]
Simplifying this:
\[ \frac{\lambda_{max}}{\lambda_{min}} = \frac{400}{9R} \cdot \frac{1}{16R} = \frac{400}{144R^2} = \frac{25}{9} \]

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