Calculate the ratio of shortest wavelength possible in Lyman and Balmer series.

Answer

To calculate the ratio of the shortest wavelength possible in the Lyman and Balmer series, we will follow these steps:

### Step 1: Understand the Series
The Lyman series corresponds to transitions where the electron falls to the n=1 energy level, while the Balmer series corresponds to transitions where the electron falls to the n=2 energy level.

### Step 2: Use the Rydberg Formula
The Rydberg formula for the wavelength of emitted light during electronic transitions 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 higher energy level.

### Step 3: Calculate Wavelength for Lyman Series
For the Lyman series:
- The lower level \( n_1 = 1 \)
- The upper level \( n_2 \) can be taken as \( \infty \) for the shortest wavelength.

Substituting these values into the Rydberg formula:
\[ \frac{1}{\lambda_1} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R \left( 1 - 0 \right) = R \]
Thus, the wavelength for the Lyman series is:
\[ \lambda_1 = \frac{1}{R} \]

### Step 4: Calculate Wavelength for Balmer Series
For the Balmer series:
- The lower level \( n_1 = 2 \)
- The upper level \( n_2 \) can also be taken as \( \infty \) for the shortest wavelength.

Substituting these values into the Rydberg formula:
\[ \frac{1}{\lambda_2} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = R \left( \frac{1}{4} - 0 \right) = \frac{R}{4} \]
Thus, the wavelength for the Balmer series is:
\[ \lambda_2 = \frac{4}{R} \]

### Step 5: Calculate the Ratio of Wavelengths
Now, we can find the ratio of the shortest wavelengths in the Lyman and Balmer series:
\[ \frac{\lambda_1}{\lambda_2} = \frac{\frac{1}{R}}{\frac{4}{R}} = \frac{1}{4} \]

### Final Answer
The ratio of the shortest wavelength possible in the Lyman series to that in the Balmer series is:
\[ \frac{\lambda_1}{\lambda_2} = \frac{1}{4} \]

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