If the wavelength of first member of Balmer series of hydrogen spectrum is `6564 A^(@)`. The wavelength of second member of Balmer series will be :

Options

A. `1215Å`
B. `4862 Å`
C. 6050
D. data given is insufficient to calculate the value

Answer

To find the wavelength of the second member of the Balmer series for the hydrogen spectrum, we can follow these steps:

### Step 1: Understand the Balmer Series Formula
The formula for the wavelength of the Balmer series is given by:

\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]

Where:
- \( \lambda \) is the wavelength,
- \( R \) is the Rydberg constant,
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)),
- \( n_f \) is the final energy level (for Balmer series, \( n_f = 2 \)),
- \( n_i \) is the initial energy level.

### Step 2: Calculate for the First Member of the Balmer Series
For the first member of the Balmer series, \( n_i = 3 \):

\[ \frac{1}{\lambda_1} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) \]

Calculating the right-hand side:

\[ \frac{1}{4} = \frac{9}{36}, \quad \frac{1}{9} = \frac{4}{36} \]

Thus,

\[ \frac{1}{\lambda_1} = R \left( \frac{9 - 4}{36} \right) = R \left( \frac{5}{36} \right) \]

### Step 3: Calculate for the Second Member of the Balmer Series
For the second member of the Balmer series, \( n_i = 4 \):

\[ \frac{1}{\lambda_2} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{4} - \frac{1}{16} \right) \]

Calculating the right-hand side:

\[ \frac{1}{4} = \frac{4}{16}, \quad \frac{1}{16} = \frac{1}{16} \]

Thus,

\[ \frac{1}{\lambda_2} = R \left( \frac{4 - 1}{16} \right) = R \left( \frac{3}{16} \right) \]

### Step 4: Relate the Two Wavelengths
Now we can relate \( \lambda_1 \) and \( \lambda_2 \):

\[ \frac{\lambda_2}{\lambda_1} = \frac{R \left( \frac{5}{36} \right)}{R \left( \frac{3}{16} \right)} = \frac{5}{36} \cdot \frac{16}{3} = \frac{80}{108} = \frac{20}{27} \]

### Step 5: Calculate \( \lambda_2 \)
Given that \( \lambda_1 = 6564 \, \text{Å} \):

\[ \frac{\lambda_2}{6564} = \frac{20}{27} \]

Thus,

\[ \lambda_2 = 6564 \cdot \frac{20}{27} \]

Calculating \( \lambda_2 \):

\[ \lambda_2 = \frac{6564 \cdot 20}{27} = 4862 \, \text{Å} \]

### Conclusion
The wavelength of the second member of the Balmer series is \( 4862 \, \text{Å} \).

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