What is the ratio of the shortest wavelength of the balmer to the shoretst of the lyman series ?

Options

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

Explanation

To find the ratio of the shortest wavelength of the Balmer series to the shortest wavelength of the Lyman series, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula**: The wavelength of emitted light in the hydrogen atom can be calculated using the formula:
\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]
where \( R \) is the Rydberg constant, \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)), \( n_f \) is the final energy level, and \( n_i \) is the initial energy level.

2. **Identify the Series**:
- For the **Balmer series**, the final energy level \( n_f = 2 \).
- For the **Lyman series**, the final energy level \( n_f = 1 \).

3. **Determine the Shortest Wavelength**:
- The shortest wavelength occurs when the initial energy level \( n_i \) approaches infinity (\( n_i \to \infty \)).
- For the **Balmer series**:
\[ \frac{1}{\lambda_{Balmer}} = RZ^2 \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = RZ^2 \left( \frac{1}{4} - 0 \right) = \frac{RZ^2}{4} \]
- For the **Lyman series**:
\[ \frac{1}{\lambda_{Lyman}} = RZ^2 \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = RZ^2 \left( 1 - 0 \right) = RZ^2 \]

4. **Calculate the Wavelengths**:
- Thus, we have:
\[ \lambda_{Balmer} = \frac{4}{RZ^2} \]
\[ \lambda_{Lyman} = \frac{1}{RZ^2} \]

5. **Find the Ratio**:
- Now, we can find the ratio of the shortest wavelength of the Balmer series to that of the Lyman series:
\[ \frac{\lambda_{Balmer}}{\lambda_{Lyman}} = \frac{\frac{4}{RZ^2}}{\frac{1}{RZ^2}} = \frac{4}{1} = 4 \]

### Final Answer:
The ratio of the shortest wavelength of the Balmer series to the shortest wavelength of the Lyman series is \( 4:1 \).