The ratio of minimum wavelengths of Lyman and Balmer series will be

Options

A. 1.25
B. 0.25
C. 5
D. 10

Answer

To find the ratio of the minimum wavelengths of the Lyman and Balmer series, we can follow these steps:

### Step 1: Understand the Concept of Minimum Wavelength
The minimum wavelength corresponds to the transition of an electron from a higher energy level (n2) to the lowest energy level (n1). For the Lyman series, the electron transitions to n1 = 1, and for the Balmer series, it transitions to n1 = 2.

### Step 2: Use the Rydberg Formula
The Rydberg formula for the wavelength (λ) of 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, \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.

### Step 3: Calculate Minimum Wavelength for Lyman Series
For the Lyman series:
- \( n_1 = 1 \)
- \( n_2 = \infty \)

Substituting these values into the Rydberg formula:
\[ \frac{1}{\lambda_L} = R_H \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R_H \left( 1 - 0 \right) = R_H \]
Thus,
\[ \lambda_L = \frac{1}{R_H} \]

### Step 4: Calculate Minimum Wavelength for Balmer Series
For the Balmer series:
- \( n_1 = 2 \)
- \( n_2 = \infty \)

Substituting these values into the Rydberg formula:
\[ \frac{1}{\lambda_B} = R_H \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = R_H \left( \frac{1}{4} - 0 \right) = \frac{R_H}{4} \]
Thus,
\[ \lambda_B = \frac{4}{R_H} \]

### Step 5: Find the Ratio of Minimum Wavelengths
Now, we need to find the ratio of the minimum wavelengths of the Lyman series to the Balmer series:
\[ \frac{\lambda_L}{\lambda_B} = \frac{\frac{1}{R_H}}{\frac{4}{R_H}} = \frac{1}{4} \]

### Step 6: Convert to Decimal
The ratio \(\frac{1}{4}\) can be expressed in decimal form:
\[ \frac{1}{4} = 0.25 \]

### Conclusion
The ratio of the minimum wavelengths of the Lyman and Balmer series is:
\[ \text{Answer: } 0.25 \]