The Wavelength of first member of Balmer series in hydrogen spectrum is `lambda` . Calculate the wavelength of first member of Lymen series in the same spectrum

Options

A. `(5//27)lambda`
B. `(4//27)lambda`
C. `(27//5)lambda`
D. `(27//4)lambda`

Answer

To solve the problem, we need to calculate the wavelength of the first member of the Lyman series in the hydrogen spectrum, given that the wavelength of the first member of the Balmer series is denoted as \( \lambda \).

### Step-by-Step Solution:

1. **Understanding the Balmer Series**:
- The Balmer series corresponds to transitions where the electron falls to the second energy level (n=2) from higher levels (n=3, 4, 5, ...).
- The first member of the Balmer series corresponds to the transition from n=3 to n=2.

2. **Using the Rydberg Formula for Balmer Series**:
- The Rydberg formula for the wavelength of emitted light is given by:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
- For the first member of the Balmer series (n=3 to n=2):
- \( n_1 = 2 \)
- \( n_2 = 3 \)
- Plugging in the values:
\[ \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) \]
- Finding a common denominator (36):
\[ \frac{1}{\lambda} = R \left( \frac{9}{36} - \frac{4}{36} \right) = R \left( \frac{5}{36} \right) \]
- Therefore, we can express this as:
\[ \frac{1}{\lambda} = \frac{5R}{36} \quad \text{(Equation 1)} \]

3. **Understanding the Lyman Series**:
- The Lyman series corresponds to transitions where the electron falls to the first energy level (n=1) from higher levels (n=2, 3, ...).
- The first member of the Lyman series corresponds to the transition from n=2 to n=1.

4. **Using the Rydberg Formula for Lyman Series**:
- For the first member of the Lyman series (n=2 to n=1):
- \( n_1 = 1 \)
- \( n_2 = 2 \)
- Plugging in the values:
\[ \frac{1}{\lambda'} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) \]
- Finding a common denominator:
\[ \frac{1}{\lambda'} = R \left( \frac{4}{4} - \frac{1}{4} \right) = R \left( \frac{3}{4} \right) \]
- Therefore, we can express this as:
\[ \frac{1}{\lambda'} = \frac{3R}{4} \quad \text{(Equation 2)} \]

5. **Finding the Relationship Between \( \lambda \) and \( \lambda' \)**:
- Now, we can relate \( \lambda \) and \( \lambda' \) by dividing Equation 1 by Equation 2:
\[ \frac{\frac{1}{\lambda}}{\frac{1}{\lambda'}} = \frac{\frac{5R}{36}}{\frac{3R}{4}} \]
- Simplifying this gives:
\[ \frac{\lambda'}{\lambda} = \frac{5R}{36} \cdot \frac{4}{3R} = \frac{5 \cdot 4}{36 \cdot 3} = \frac{20}{108} = \frac{5}{27} \]
- Therefore, we have:
\[ \lambda' = \frac{5}{27} \lambda \]

### Final Answer:
The wavelength of the first member of the Lyman series in the hydrogen spectrum is:
\[ \lambda' = \frac{5}{27} \lambda \]