What is the ratio of the shortest wavelength of the balmer to the shortest 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 1: Understand the Series
The Lyman series corresponds to transitions where the electron falls to the n=1 level, while the Balmer series corresponds to transitions where the electron falls to the n=2 level.

### Step 2: Use the Rydberg Formula
The Rydberg formula for the wavelength of the emitted light is given by:

\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

where \( R \) is the Rydberg constant, \( n_1 \) is the lower energy level, and \( n_2 \) is the upper energy level.

### Step 3: Calculate the Shortest Wavelength for the Balmer Series
For the Balmer series, the transition is from \( n_2 \) (any level greater than 2) to \( n_1 = 2 \). The shortest wavelength corresponds to the transition from \( n_2 = \infty \) to \( n_1 = 2 \):

\[ \frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = R \left( \frac{1}{4} - 0 \right) = \frac{R}{4} \]

Thus, the wavelength for the Balmer series is:

\[ \lambda_B = \frac{4}{R} \]

### Step 4: Calculate the Shortest Wavelength for the Lyman Series
For the Lyman series, the transition is from \( n_2 \) (any level greater than 1) to \( n_1 = 1 \). The shortest wavelength corresponds to the transition from \( n_2 = \infty \) to \( n_1 = 1 \):

\[ \frac{1}{\lambda_L} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R \left( 1 - 0 \right) = R \]

Thus, the wavelength for the Lyman series is:

\[ \lambda_L = \frac{1}{R} \]

### Step 5: Find the Ratio of the Wavelengths
Now we can find the ratio of the shortest wavelength of the Balmer series to the shortest wavelength of the Lyman series:

\[ \text{Ratio} = \frac{\lambda_B}{\lambda_L} = \frac{\frac{4}{R}}{\frac{1}{R}} = \frac{4}{1} = 4 \]

### Final Answer
The ratio of the shortest wavelength of the Balmer series to the shortest wavelength of the Lyman series is:

\[ \text{Ratio} = 4 \]

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