The ratio of minimum to maximum wavelength in Balmer series is

Options

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

Answer

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 Formula
The Balmer series describes the spectral lines of hydrogen when an electron transitions from a higher energy level (n ≥ 3) to the second energy level (n = 2). The formula for the wavelength (λ) in the Balmer series is given by:

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

where \( R \) is the Rydberg constant, and \( n \) is the principal quantum number of the higher energy level.

### Step 2: Calculate Maximum Wavelength (λ_max)
For the maximum wavelength (λ_max), the transition occurs from n = 3 to n = 2. Plugging in the values:

\[ \frac{1}{\lambda_{max}} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]

Calculating this:

\[ \frac{1}{\lambda_{max}} = R \left( \frac{1}{4} - \frac{1}{9} \right) \]

Finding a common denominator (36):

\[ \frac{1}{\lambda_{max}} = R \left( \frac{9}{36} - \frac{4}{36} \right) = R \left( \frac{5}{36} \right) \]

Thus,

\[ \lambda_{max} = \frac{36}{5R} \]

### Step 3: Calculate Minimum Wavelength (λ_min)
For the minimum wavelength (λ_min), the transition occurs from n = ∞ to n = 2. Plugging in the values:

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

Since \( \frac{1}{\infty^2} = 0 \):

\[ \frac{1}{\lambda_{min}} = R \left( \frac{1}{4} \right) \]

Thus,

\[ \lambda_{min} = \frac{4}{R} \]

### Step 4: Calculate the Ratio of Minimum to Maximum Wavelength
Now, we can find the ratio of λ_min to λ_max:

\[ \text{Ratio} = \frac{\lambda_{min}}{\lambda_{max}} = \frac{\frac{4}{R}}{\frac{36}{5R}} = \frac{4}{R} \cdot \frac{5R}{36} = \frac{20}{36} = \frac{5}{9} \]

### Conclusion
The ratio of the minimum to maximum wavelength in the Balmer series is \( \frac{5}{9} \).