The ratio of minimum to maximum wavelength in Balmer series is

Options

A. `5:9`
B. `5:36`
C. `1:4`
D. `3:4`

Explanation

To find the ratio of the minimum to maximum wavelength in the Balmer series, we can follow these steps:

### Step 1: Understand the Balmer Series
The Balmer series corresponds to the transitions of electrons in a hydrogen atom where the final energy level (n1) is 2. The initial energy level (n2) can be any integer greater than 2 (n2 = 3, 4, 5, ...).

### Step 2: Use the Rydberg Formula
The Rydberg formula for the wavelength (λ) of the emitted light is given by:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R_H \) is the Rydberg constant.

### Step 3: Calculate the Minimum Wavelength (λ_min)
For the minimum wavelength, the transition occurs from n2 = ∞ to n1 = 2:
\[ \frac{1}{\lambda_{\text{min}}} = R_H \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) \]
Since \( \frac{1}{\infty^2} = 0 \):
\[ \frac{1}{\lambda_{\text{min}}} = R_H \left( \frac{1}{4} \right) \]
Thus,
\[ \lambda_{\text{min}} = \frac{4}{R_H} \]

### Step 4: Calculate the Maximum Wavelength (λ_max)
For the maximum wavelength, the transition occurs from n2 = 3 to n1 = 2:
\[ \frac{1}{\lambda_{\text{max}}} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
Calculating the right-hand side:
\[ \frac{1}{\lambda_{\text{max}}} = R_H \left( \frac{1}{4} - \frac{1}{9} \right) \]
Finding a common denominator (36):
\[ \frac{1}{\lambda_{\text{max}}} = R_H \left( \frac{9 - 4}{36} \right) = R_H \left( \frac{5}{36} \right) \]
Thus,
\[ \lambda_{\text{max}} = \frac{36}{5R_H} \]

### Step 5: Find the Ratio of λ_min to λ_max
Now we can find the ratio of the minimum wavelength to the maximum wavelength:
\[ \text{Ratio} = \frac{\lambda_{\text{min}}}{\lambda_{\text{max}}} = \frac{\frac{4}{R_H}}{\frac{36}{5R_H}} = \frac{4}{R_H} \times \frac{5R_H}{36} = \frac{4 \times 5}{36} = \frac{20}{36} = \frac{5}{9} \]

### Final Answer
The ratio of the minimum to maximum wavelength in the Balmer series is:
\[ \frac{5}{9} \]