What is the ratio of the wave lengths of last lines of Balmer and Lyman series

Answer

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

### Step 1: Identify the transitions for the last lines of the series
- The last line of the Balmer series corresponds to the transition from \( n = \infty \) to \( n = 2 \).
- The last line of the Lyman series corresponds to the transition from \( n = \infty \) to \( n = 1 \).

### Step 2: Use the Rydberg formula for the Balmer series
The Rydberg formula for the wavelength (\( \lambda \)) of the emitted light during a transition is given by:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
For the Balmer series (last line):
- \( n_1 = 2 \)
- \( n_2 = \infty \)

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

### Step 3: Use the Rydberg formula for the Lyman series
For the Lyman series (last line):
- \( n_1 = 1 \)
- \( n_2 = \infty \)

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

### Step 4: Calculate the ratio of the wavelengths
Now we can find the ratio of the wavelengths of the last lines of the Balmer and Lyman series:
\[ \frac{\lambda_B}{\lambda_L} = \frac{\frac{4}{R}}{\frac{1}{R}} = 4 \]

### Final Answer
The ratio of the wavelengths of the last lines of the Balmer and Lyman series is \( 4:1 \).

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